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ST TEE SAME AUTHOR, 



AN ELEMENTARY TREATISE ON ANALYTIC GEOMETRY, 



EMBEAC1NG 



PLANE GEOMETRY, 



INTRODUCTION TO GEOMETRY OF THREE DIMENSIONS. 



AN 



ELEMENTARY TREATISE 



DIFFERENTIAL AND INTEGRAL 



CALCULUS 



WITH NUMEROUS EXAMPLES. 



BY 

EDWARD A: BOWSER. 

PBOFESSOB OF MA THEMATIC? AND ENGINEEBING IN RUTGEE3 COLLEGE. 



NEW YORK: 

D. VAN NOSTRAND, 

23 Muebay Street. 
1880. 



QK^o3 
&S8 



COPYRIGHT, 1880, by E. A. BOWSER. 



PREFACE 



rpHE present work on the Differential and Integral Calculus is 
designed as a text-book for colleges and scientific schools. The 
aim has been to exhibit the subject in as concise and simple a manner 
as was consistent with rigor of demonstration, to make it as attractive 
to the beginner as the nature of the Calculus would permit, and to 
arrange the successive portions of the subject in the order best suited 
for the student. 

I have adopted the method of infinitesimals, having learned from 
experience that the fundamental principles of the subject are made 
more intelligible to beginners by the method of infinitesimals than by 
that of limits, while in the practical applications of the Calculus the 
investigations are carried on entirely by the method of infinitesimals. 
At the same time, a thorough knowledge of the subject requires that 
the student should become acquainted with both methods ; and for 
this reason, Chapter III is devoted exclusively to the method of 
limits. In this chapter, all the fundamental rules for differentiating 
algebraic and transcendental functions are obtained by the method of 
limits, so that the student may compare the two methods. This chap- 
ter may be omitted without interfering with the continuity of the 
work, but the omission of at (east the first part of the chapter is not 
recommended. 

To familiarize the student with the principles of the subject, and 
to fix the principles in his mind, a large number of examples is given 
at the ends of the chapters. These examples have been carefully 
selected with the view of illustrating the most important points of 
the subject. The greater part of them will present no serious diffi- 
culty to the student, while a few may require some analytical skill. 



IV PREFACE. 

In preparing this book, I have availed myself pretty freely of the 
writings of the best American and English and French authors. 
Many volumes have been consulted whose titles are not mentioned, 
as credit could not be given in every case, and probably I am indebted 
to these volumes for more than I am aware of. The chief sources 
upon which I have drawn are indicated by the references in the body 
of the work, and need not be here repeated. For examples, I have 
drawn upon the treatises of Gregory, Price, Todhunter, Williamson, 
Young, Hall, Rice and Johnson, Ray, and Olney, while quite a num- 
ber has been taken from the works of De Morgan, Lacroix, Serret, 
Courtenay, Loomis, Church, Byerly, Docharty, Strong, Smyth, and 
the Mathematical Visitor; and I would hereby acknowledge my 
indebtedness to all the above-named works, both American and 
foreign, for many valuable hints, as well as for examples. A few 
examples have been prepared specially for this work. 

I have again to express my thanks to Mr. R. W. Prentiss, Fellow 
in Mathematics at the Johns Hopkins University, for reading the MS, 
and for valuable suggestions. 

B. A. B. 
Rutgers College, | 

New Brunswick, N. J., June, 1880. \ 



TABLE OF CONTENTS 



PART I. 
DIFFERENTIAL CALCULUS 



CHAPTER I. 

FIRST PRINCIPLES. 

AKT. PAGE 

1 . Constants and Variables 1 

2. Independent and Dependent Variables 1 

3. Functions. Geometric Representation 2 

4. Algebraic and Transcendental Functions 4 

5. Increasing and Decreasing Functions 5 

6. Explicit and Implicit Functions 6 

7. Continuous Functions 6 

8. Infinites and Infinitesimals 7 

9. Orders of Infinites and Infinitesimals, 8 

10. Geometric Illustration of Infinitesimals 10 

11 . Axioms 12 

Examples 13 



CHAPTER II. 

DIFFERENTIATION OF ALGEBRAIC AND TRANSCENDENTAL 
FUNCTIONS. 

12. Increments and Differentials 15 

13. Consecutive Values 16 

14. Differentiation of Sum of a Number of Functions 16 

15. To Differentiate y = ax ± b 17 



VI CONTENTS. 

ABT. PAGE 

16 Differentiation of a Product of Two Functions 18 

17. Differentiation of a Product 20 

18. Differentiation of a Fraction 20 

19. Differentiation of any Power 21 

Examples 23 

Illustrative Examples 26 

20. Logarithmic and Exponential Functions 29 

21 . Differentiation of an Exponential 31 

22. Differentiation of an Exponential with Variable Base 32 

Examples 33 

23. Logarithmic Differentiation. Examples 34 

Illustrative Examples „ 35 

TRIGONOMETRIC FUNCTIONS. 

24. To Differentiate y = sin x 37 

25. To Differentiate y = cos x 38 

28. To Differentiate y = tan x 88 

27. To Differentiate y = cot x . 39 

23. To Differentiate y = sec a? « 39 

29. To Differentiate y — cosec x 40 

30. To Differentiate y = versin x 40 

31. To Differentiate y = covers x 40 

32. Geometric Demonstration 41 

Examples 43 

Illustrative Examples 44 



CIRCULAR FUNCTIONS. 

33. To Differentiate y — sin" 1 x 46 

34. To Differentiate y = cos- 1 x 46 

85. To Differentiate y = tan- 1 x 47 

36. To Differentiate y = cot" 1 x 47 

87. To Differentiate y — sec- 1 x 47 

38. To Differentiate y — cosec- 1 x 47 

80. To Differentiate y = vers-' x 48 

40. To Differentiate y = covers -1 x 48 

Examples . 48 

Miscellaneous Examples 49 



CONTENTS. Vll 



CHAPTEE III. 

LIMITS — DERIVED FUNCTIONS. 

ART. PAGE 

41. Limiting Values 59 

42. Algebraic Illustration 59 

43. Trigonometric Illustration 60 

44. Derivatives 62 

45. Differential and Differential Coefficient 63 

46. Algebraic Sum of a Number of Functions C3 

47. Product of Two Functions 64 

48. Product of any Number of Functions 65 

49. Differentiation of a Fraction QQ 

50. Any Power of a Single Variable 67 

51. Differentiation of log x 68 

52. Differentiation of a* 68 

53. Differentiation of sin x 63 

54. Differentiation of cos x 69 



CHAPTER IV. 

SUCCESSIVE DIFFERENTIALS AND DERIVATIVES. 

55. Successive Differentials. Examples , 71 

56. Successive Derivatives 74 

56a. Geometric Representation of First Derivative. Examples . 76 

CHAPTER V. 

DEVELOPMENT OF FUNCTIONS. 

57. Definition of Development of a Function 81 

58. Maclaurin's Theorem 81 

59 The Binomial Theorem 85 

60. To Develop y = sin x and y — cos x 85 

61. The Logarithmic Series , 88 

62. The Exponential Series 90 

63. To Develop y = tan- 1 x 91 

64. Failure of Maclaurin's Theorem 92 

65. Taylor's Theorem. Lemma 93 

ffi. To find Taylor's Theorem 95 

67. The Binomial Theorem 97 



Till • CONTENTS, 

***' PAGE 

68. To Develop u' = sin {x + y) 98 

69. The Logarithmic Series 93 

70. The Exponential Series 93 

71. Failure of Taylor's Theorem 99 

Examples. iqq 



CHAPTER VI. 

EVALUATION OF INDETERMINATE FORMS. 

72. Indeterminate Forms 103 

78. Common Factors. Examples , 104 

74. Method of the Differential Calculus 105 

75. To evaluate Functions of the form — 108 

00 

78. To evaluate Functions of the form Oxoo Ill 

77. To evaluate Functions of the form 00 — 00 112 

78. To evaluate Functions of the forms 0°, 00 °, and 1 ±Q0 113 

79. Compound Indeterminate Forms 116 

Examples 116 



CHAPTER VII. 

FUNCTIONS OF TWO OR MORE VARIABLES. — CHANGE OF 
THE INDEPENDENT VARIABLE. 

80. Partial Differentiation ISO 

81. Differentiation of a Function of Two Variables 122 

82. To find the Total Derivative of u with respect to x , 125 

83. Successive Partial Differentiation 130 

84. Proof that Order of Differentiation is indifferent 131 

85 Successive Differentials of a Function of Two Independent 

Variables 133 

86. Implicit Functions 135 

87. Differentiation of an Implicit Function 133 

88. Second Derivative of an Implicit Function 138 

89. Change of the Independent Variable 140 

90. General Values of %, f-^', ~|, etc 141 

dx dx- dx z 

91. Transformation for Two Independent Variables 145 

Examples , 147 



CONTEXTS. IX 



CHAPTER VIII. 

MAXIMA AND MINIMA OF FUNCTIONS OF A SINGLE 

VARIABLE. 

jfcBT. PAGE 

02. Definition of a Maximum and a Minimum 151 

93. Condition for a Maximum or Minimum 151 

94. Geometric Illustration 152 

95. Discrimination between Maxima and Minima 154 

96. Condition given by Taylor's Theorem 154 

97. Method of finding Maxima and Minima Values 155 

98. Alternation of Maxima and Minima Values 156 

99. Application of Axiomatic Principles. 157 

Examples 159 

Geometric Problems 164 

CHAPTER IX. 

TANGENTS, NORMALS, AND ASYMPTOTES. 

100. Equations of the Tangent and Normal 172 

101. Length of Tangent, Normal, Subtangent, etc 175 

102. Polar Curves. Tangents, Normals, Subtangents, etc 178 

103. Rectilinear Asymptotes 181 

104. Asymptotes determined by Expansion 184 

105. Asymptotes in Polar Co-ordinates. Examples 186 

CHAPTER X. 

DIRECTION OF CURVATURE — SINGULAR POINTS — TRACING 
OF CURVES. 

106. Concavity and Convexity 191 

107. Polar Co-ordinates 192 

108. Singular Points '. 194 

109. Points of Inflexion 194 

110. Multiple Points ... 106 

111. Cusps 199 

112. Conjugate Points 201 

113. Shooting Points. Stop Points 203 

114. Tracing Curves 205 

Examples 206 

115. Tracing Polar Curves. Examples 210 



CONTENTS. 



CHAPTER XI. 

RADIUS OF CURVATURE, EVOLUTES AND INVOLUTES — 
ENVELOPES. 

ART. PAGE 

116. Curvature 216 

117. Order of Contact of Curves 217 

118. Dependence of Order of Contact on Arbitrary Constants 218 

119. Radius of Curvature. Centre of Curvature 219 

120. Second Method 220 

121. Radius of Curvature in Polar Co-ordinates. 222 

122. Radius of Curvature at a Maximum or Minimum. 222 

123. Contact of Different Orders 223 

Examples 224 

124. Evolutes and Involutes 223 

125. Equation of the Evolute 228 

126. Normal to an Involute is tangent to Evolute 230 

127. Envelopes of Curves 231 

128. Equation of the Envelope of a Series of Curves 232 

Examples 233 



PART II. 
INTEGRAL CALCULUS 



CHAPTER I. 

ELEMENTARY FORMS OF INTEGRATION. 

129. Definitions ' 238 

130. Elementary Rules for Integration 2G9 

131. Fundamental Forms 242 

132. Integration by Transformation into Fundamental Forms 243 

loo. Integrating Factor. Examples 217 

134, Transposition of Variable Factors. Examples 2 19 

135. Trigonometric Reduction. Examples 254 



CONTENTS. xi 



CHAPTER II. 

INTEGRATION OF RATIONAL FRACTIONS. 

ART. PAGE 

136. Rational Fractions 256 

137. Case 1. Decomposition of a Rational Fraction 256 

138. Case 2. " " " 259 

139. Case 3. " * " 262 

Examples 263 

CHAPTER III. 

INTEGRATION OF IRRATIONAL FUNCTIONS BY 
RATIONALIZATION. 

140. Rationalization 269 

141. Monomial Surds 269 

142. Binomial Surds of the First Degree 270 

143. Functions of the Form — t 272 

(a + bx 2 ) 2 

144. Functions containing only Trinomial Surds 272 

145. Binomial Differentials 276 

v_ 

146. Conditions for Rationalization of x m (a + bx n )i dx 277 

Examples 280 

CHAPTER IV. 

INTEGRATION BY SUCCESSIVE REDUCTIONS. 

147. Formulae of Reduction 285 

148. Formula for Diminishing Exponent of x, etc 285 

149. Formula for Increasing Exponent of a;, etc 287 

150. Formula for Diminishing Exponent of Parenthesis 288 

151. Formula for Increasing Exponent of Parenthesis 289 

Examples. Applications of Formulae 289 

Logarithmic Functions 295 

152. Reduction of the Form / X(log x) n dx 295 

/x m dx 

i — -- , 297 
log" x 



. a T dx 
155. Redaction of the Form / — - 300 



XI l CONTENTS. 

ART. PAGE 

Exponential Forms 299 

154. Reduction of the Form / a mx x n dx 299 

/oh 

156. Trigonometric Functions 301 

157. Formulae of Reduction for / sin w 6 cos" 6 dB 303 

158. Integration of sin" 1 6 cos 11 Odd 305 

159. Reduction of the Form / x n cos ax dx 307 

160. Reduction of the Form / e ax cos" x dx 308 

161. Integration of f(x) sin- 1 x dx, f(x) tan- 1 x dx, etc 309 

Hft 

162. Integration of dy = 310 

a + 6 cos 6 

Examples • 312 

CHAPTEE V. 

INTEGRATION BY SERIES — SUCCESSIVE INTEGRATION — IN- 
TEGRATION OF FUNCTIONS OF TWO VARIABLES. 
DEFINITE INTEGRALS. 

163. Integration by Series 319 

164. Successive Integration 321 



/ 



165. 



/n 

X dx H into a Series 323 

166. Integrations of Functions of Two or More Variables 326 

167. Integration of -^_ =f(x,y) 326 

(ty ciiX 

168. Integration of Total Differentials of the First Order 329 

169. Definite Integrals. Examples 331 

170. Change of Limits 334 

Examples 337 

Formulae of Integration 340 

CHAPTER VI. 

LENGTHS OF CURVES. 

171. Length of Plane Curves referred to Rectangular Axes 347 

172. Rectification of Parabola , 348 



CONTEXTS. Xil! 

ART. PAGE 

173. Semi-cubical Parabola 349 

174. The Circle 349 

175. The Ellipse 350 

176. The Cycloid 351 

177. The Catenary 352 

178. The Involute of a Circle 352 

179. Rectification in Polar Co-ordinates 353 

180. The Spiral of Archimedes 353 

181. The Cardioide 353 

182. Length of Curves in Space : 354 

183. Intersection of Two Cylinders. Examples 355 

CHAPTER VII. 

AREAS OF PL AXE CURVES. 

184. Areas of Curves 360 

185. Area between Two Curves 361 

186. Area of the Circle 361 

187. The Parabola 362 

188. The Cycloid 362 

189. The Ellipse : 363 

190. Area between Parabola and Circle 363 

191. Area in Polar Co-ordinates 364 

192. The Spiral of Archimedes 364 

Examples 365 

CHAPTER VIII. 

AREAS OF CURVED SURFACES. 

193. Surfaces of Revolution 369 

194. Quadrature of the Sphere 370 

195. The Paraboloid of Revolution 371 

196. The Prolate Spheroid 372 

197. The Catenary 372 

198. The Surface of Revolution generated by Cycloid 373 

199. Surface of Revolution in Polar Co-ordinates 374 

200. The Cardioide 374 

201. Any Curved Surfaces. Double Integration 375 

202. Surface of the Octant of a Sphere 376 

Examples 377 



XIV CONTENTS. 



CHAPTEE IX. 

VOLUMES OF SOLIDS. 

ART. PAGE 

203. Solids of Revolution 381 

204 The Sphere 381 

205. Solid of Revolution of Cycloid 382 

206. Solid of Revolution generated by Cissoid 383 

207. Volumes of Solids bounded by any Curved Surface 383 

208. Mixed System of Co-ordinates 388 

209. Cubature in Polar Co-ordinates 389 

Examples 390 



PART I. 

DIFFERENTIAL CALCULUS 



CHAPTER I. 

FIRST PRINCIPLES. 

1. Constants and Variables— In the Calculus, as in 
Analytic Geometry, there are two kinds of quantities used, 
constants and variables. 

A constant quantity, or simply a constant, is one whose 
yalue does not change in the same discussion, and is repre- 
sented by one of the leading letters of the alphabet. 

A variable quantity, or simply a variable, is one which 
admits of an infinite number of values within certain limits 
that are determined by the nature of the problem, and is 
represented bv one of the final letters of the alphabet. 

For example, in the equation of the parabola, 

if = 2px, 

x and y ar-3 variables, as they represent the co-ordinates of 
any point of the parabola, and so may have an indefinite 

of different values. 2p is a constant, as it represents 
th bus rectum of the parabola, and so has but one fixed 

\.ny given number is constant. 

2. Independent and Dependent Variables. — An 

■ a lent variable is one to which any arbitrary value may 



2 FUNCTION OF ONE OR MORE VARIABLES. 

be assigned at pleasure. A dependent variable is one whose 
value varies in consequence of the variation of the inde- 
pendent variable or variables with which it is connected. 
Thus, in the equation of the circle 

■x 2 + y 2 = r 2 , 

if wt? assign to x any arbitrary value, and find the correspond- 
ing value of y, we make x the independent variable, and y 
the dependent variable. If* we were to assign to y any arbi- 
trary value, and £nd the corresponding value of x, we would 
make y the independent variable and x the dependent 
variable. 

Frequently, when we are considering two or more varia- 
bles, it is in our power to make whichever we please the 
independent variable. But, having once chosen the inde- 
pendent variable, we are not at liberty to change it through- 
out our operations, unless we mak<P the corresponding trans- 
formations which such a change would require. 

3. Functions.— One quantity is called a function of 
another, when it is so connected with itk that no change can 
take place in the latter without producing a corresponding 
change in the former. 

For example, the sine, cosine, tangent, etc., of an angle 
are said to be functions of the angle, as they depend upon 
the angle for their value. Also, the area of a square is a 
function of its side; the volume of a sphere is sa function of 
its radius. In like manner, any algebraic expression in x, as 

x B — 2bx 2 + bx -f c, 

is a function of x. Also, we may have a function of two or 
more variables: a rectangle is a function of its t\vo sid 
a parallelopiped is a function of its three edges ; tl ^ 
sion tan (ax + by) is a function of two variables, x anil y\ 
tf' 4- y 2 + z 2 is a function of three variables, x, y, anri z ; etc. 
When we wish to write that one quantity is a function of 



NOTATION — GEOMETRIC REPRESENTATION. 3 

one or more others, and wish, at the same time, to indicate 
several forms of functions in the same discussion, we use 
such symbols as the following : 

V - f{*) ; V = F{x) ; y = (x) ; y =/'(«); 

y == J\*> ■*) ; *<*, jr) == j / te y> «) = ; 

which are read: "?/ equals the /function of x\ y eqnals the 
large F function of x ; y equals the function of x ; y equals 
the /prime function of x; y equals the /function of x and 
z; the function of x and y equals zero ; the / function of 
x, y, and z equals zero;" or sometimes " y = / of %, 
y = F of x," etc. If we do not care to state precisely the 
form of the function, we may read the above, "y = a func- 
tion of x ; ?/ = a function of a; and z ; a function of 2; and y 
= ; a function of ar, #, and z = 0." 
For example, in the equation 

# = fliC 3 + foe -f- c, 

?/ is a function of #, and may be expressed, y =f(x). 
Also, the equation 

ax- 4- £.i^ 4- c?/ = 

may be expressed, f(x, y) = 0. 
In like manner, the equations 

y = ax z 4- bx 2 z 4- cz 3 , 

and • ?/ = ax 2 4- te 4- dz 2 , 

may be expressed, # =f(x, z) and y = $(x, z). 

Every function of a single variable may be represented geometri- 
cally by the ordinate of a curve of which the variable is the cor- 
responding abscissa. For if y be any function of x, and we assign 
any value to x and find the corresponding value of y, these two values 
may be regarded as the co-ordinates of a point which may be con- 
structed. In the same way, any number of values maybe assigned to 
<r, and the corresponding values of y found, and a series of points con- 



4 ALGEBRAIC AND TRANSCENDENTAL FUNCTIONS. 

structed. These points make; up a curve of which the variable ordi- 
nate is y and the corresponding abscissa is x. 

In like manner it may be shown that a function of two variables 
may be represented geometrically by the ordinate of a surface of which 
the variables are the corresponding abscissas. 

4. Algebraic and Transcendental Functions.— An 

algebraic function is one in which the only operations indi- 
cated are addition, subtraction, multiplication, division, 
involution, and evolution; as, 

(a + wyi (*-%)*; i/jLz* etc . 

V (x 2 + a z Y 

Transcendental functions are those which involve other 
operations, and are subdivided into trigonometric, circular, 
logarithmic, and exponential. 

A trigonometric function is one which involves sines, tan- 
gents, cosines, etc., as variables. For example, 

y = sin x ; y = tan 2 x ; y — cos x sec x ; etc. 

A circular function is one in which the concept is a 
variable arc, as sin -1 x,* cos _1 #, sec -1 ?/, cot" 1 a:, etc., read, 
" the arc whose sine is x, the arc whose cosine is x," etc. It 
is the inverse of the trigonometric function ; thus, from the 
trigonometric function, y = smx, we obtain the circular 
function, x = sin -1 y. In the first function we think of the 
right line, the sine, the arc being given to tell us which sine ; 
in the second we think of the arc, the sine being given to 
tell us which arc. The circular functions are often called 
inverse trigonometric functions. 

* This notation was suggested by the use of the negative exponents in algebra. 
If we have y = ax, we also have x = cr'y, where y is a function of x, and x is the 
corresponding inverse function of y. It may be worth while to caution the begin- 
ner against the error of supposing that sin" 1 y is equivalent to — — ; while it is 
true that a Ms equivalent to - • 



INCREASING AND DECREASING FUNCTIONS. 5 

A logarithmic function is one which involves logarithms 
of the variables ; as, 

y — log x ; y = log Va — x; 



/a 2 — a? 

^ = 3l0 «V^+^ ; etc- 



An exponential function is one in which the variable 
enters as an exponent ; as, 

y = a x ; y = x z ; u = x x " ; etc. 

5. Increasing and Decreasing Functions. — An in- 
creasing function is one that increases when its variable 
increases, and decreases when its variable decreases. 

For example, in the equations 



y = ax z , y = log x, y = V« 2 + # 2 , ?/ =r a 31 , 
y is an increasing function of x. 

A decreasing function is one that decreases when its 
variable increases, and increases when its variable decreases. 
Thus, in the equations 

y — -, y = (« - ^) 3 > y = % -, x 2 + y 2 = r% 

y is a decreasing function of x. In the expression, 

y = (a - z) 2 , 

y is a decreasing function for all values of x < a, but in- 
creasing for all values > a. In the expression 

y = sin .r, 

?/ is an increasing function for all values of x between 0° 
and 90°, decreasing for all values of x between 90° and 2 TO , 
and increasing for all values of x between 270° and 360°. 



6 CONTINUOUS FUNCTIONS. 

6. Explicit and Implicit Functions. — An explicit 
function is one whose value is directly expressed in terms of 
the variable and constants. 

For example,, in the equations 

y = (a — x) 2 , y = V 'a 2 — x\ y = "lax 2, — dx^ 9 

y is an explicit function of x. 

An implicit function is one whose value is not directly 
expressed in terms of the variables and constants. 

For example, in the equations 

y z — Saxy -f x B = 4, x 2 — Sxy + 2y = 16, 

y is an implicit function of x, or x is an implicit function 
of y. If we solve either equation with respect to y, we shall 
have y as an explicit function of x ; also, if we solve for x, 
we shall have x as an explicit function of y. 

7. Continuous Functions. — A function of x is said to 
be a continuous function of x, between the limits a and i, 
when, for every value of x between these limits, the cor- 
responding value of the function is finite, and when an 
infinitely small change in the value of x produces only an 
infinitely small change in the value of the function. If 
these conditions are not fulfilled, the function is discon- 
tinuous. 

For example, both conditions are fulfilled in the equations 

y = ax -f b, y = sin x, 

in which, as x changes, the value of the function also 
changes, but changes gradually as x changes gradually, and 
there is no abrupt passage from one value to another; if x 
receives a very small change, the corresponding change in 
the function of x is also very small. 

The expression yV 2 — ^ 2 is a continuous function of x 
for all values of x between + r and — r, while Vx 2 — r 2 
is discontinuous between the same limits. 



INFINITES AND INFINITESIMALS. 7 

8. Infinites and Infinitesimals. — An infinite quantity, 
or an infinite, is a quantity which is greater than any assign- 
able quantity. 

An infinitesimal is a quantity which is less than any 
assignable qu an tity. 

An infinite is not the largest possible quantity, nor is an infinitesi- 
mal the smallest ; there would, in this case, be but one infinite or 
infinitesimal. Infinites may differ from each other and from a quan- 
tity which transcends every assignable quantity, that is. from absolute 
infinity. So may infinitesimals differ from each other and from abso- 
lute zero. 

The terms infinite and infinitesimal are not applicable to quantities 
in themselves considered, but only in their relation to each other, or to 
a common standard. A magnitude which is infinitely great in com- 
parison with ^finite magnitude is said to be infinitely great. Also, a 
magnitude which is infinitely small in comparison with a finite mag- 
nitude is said to be infinitely small. Thus, the diameter of the earth 
is very great in comparison with the length of one inch, but very small 
in comparison with the distance of the earth from the pole star ; and 
it would accordingly be represented by a very large or a very small 
number, according to which of these distances is assumed as the unit 
of comparison. 

The symbols oo and are used to represent an infinite 
and an infinitesimal respectively, the relation of which is 

oo = - and = 

U oo 

The cipher is an abbreviation to denote an indefinitely small 
quantity, or an infinitesimal — that is, a quantity which is less than 
any assignable quantity— and does not mean absolute zero ; neither 
does oo express absolute infinity. 



If a represents a finite quantity, and x an infinite, then 

- is an infinitesimal. If x is an infinitesimal and a is finite, 

x 7 

a 



x 



is infinite ; that is, the reciprocal of an infinite is infini- 
tesimal, and the reciprocal of an infinitesimal is infinite. 
A number is infinitely great in comparison with another. 



8 ORDERS OF INFINITES AND INFINITESIMALS. 

when no number can be found sufficiently large to express the 

ratio betiueen them. Thus, x is infinitely great in relation 

to a. when no number can be found large enough to express 

x 
the quotient — Also, a is infinitely small in relation to x 

when no number can be found small enough to express the 

quotient -; x and - represent an infinite and an infini- 

x x 

tesimal. 

One million in comparison with one millionth is a very large num- 
ber, but not infinitely large, since the ratio of the first to the second 
can be expressed in figures : it is one trillion : though a very large 
number, it is finite. So, also, one millionth in comparison with one 
million is a very small number, but not infinitely small, since a num- 
ber can be found small enough to express the ratio of the first to the 
second : it is one trillionth, and therefore finite. 

9. Orders of Infinites and Infinitesimals. — But even 

though - is greater than any quantity to which we can 
a 

assign a value, we may suppose another quantity as large in 
relation to x as x is in relation to a : for, whatever the mag- 
nitude of x, we may have the proportion 

x* 
a : x :: x : — , 

a 

or 
m which — is as large in relation to x as x is in relation to 

a 

x 
a, for — will contain x as many times as x will contain a ; 

x 2 
hence, — may be regarded as an infinite of the second order, 

x 

- being an infinite of the first order. 

Also, even though - is less than any quantity to which 

x 

we can assign a value, we may suppose another quantity as 
small in relation to a as a is in relation to x ; for we may 
have the proportion, 



ORDERS OF INFINITES AND INFINITESIMALS. 



x : a :: a : — , 
x 

a 2 
in which — is as small in relation to a as a is in relation to 
x 

a 2 . 

x, ibr — is contained as many times in a as a is contained 



x 



ci 



in x : hence, — may be regarded as an infinitesimal of the 



second order, - being an infinitesimal of the first order. 

x 

We may, again, suppose quantities infinitely greater and 

infinitely less than these just named : and so on indefinitely. 

Thus, in the series 

c, 9 a a a 

ax 3 , ax 2 , ax, a, -, - , -„, etc., 

x x 2 x 3 ' 

if we suppose a finite and x infinite, it is clear that any 
term is infinitely small with respect to the one that imme- 
diately precedes it, and infinitely large with respect to the 
one that immediately follows it; that is, ax 3 , ax 2 , ax are 
infinites of the third, second, and first orders, respectively; 

-, — 2 , — 3 are infinitesimals of the first, second, and third 

orders, respectively, while a is finite. 

If two quantities, as x and //, are infinitesimals of the first 
order, their product is an infinitesimal of the second order ; 
for we have the proportion, 

I : x : : y : xy. 

Hence, if x is infinitely small in relation to 1, xy is infinitely 
small in relation to y; that is, it is an infinitesimal of the 
second order when x and y are infinitesimals of the first 
order. 

Likewise, the product of two infinites of. the first order is 
an infinite of the second order. 

The product of an infinite and an infinitesimal of the 
same order is a finite quantity. The product of an infinite 



10 RATIOS OF INFINITESIMALS. 

and an infinitesimal of different orders is an infinite or an 
infinitesimal, according as the order of the infinite is higher 
or lower than that of the infinitesimal, and the order of the 
product is the sum of the orders of the factors. 
For example, in the expressions 



o a 9 9 a a a A 

ax z x — o = a 2 , ax 2 x - = a 2 x, ax x — Q = -5 

X s X ' X 3 ' X 1 



the first product is finite; the second is an infinite of the 
first order ; the third is an infinitesimal of the second order. 

Though two quantities are each infinitely small, they may have any 
ratio whatever. 

Thus, if a and b are finite and x is infinite, the two quantities 

- and - are infinitesimals; but their ratio is =- , which is finite. In- 
deed, two very small quantities may have a much larger ratio than 
two very large quantities, for the value of a ratio depends on the rela- 
tive, and not on the absolute magnitude of the terms of the ratio. The 
ratio of the fraction one-millionth to one -ten-millionth is ten, while the 
ratio of one million to ten million is one-tenth. The latter numbers are 
respectively a million times a million, and ten million times ten mil- 
lion, times as great as the first, and yet the ratio of the last two is 
only one-hundredth as great as the ratio of the first two. 
Assume the series 



- f-Y (- 



( Toy ' (ioy ' (joy ' etc 



in which the first fraction is one-millionth, the second one-millionth 
of the first, and so on. Now suppose the first fraction is one-millionth 
of an inch in length, which may be regarded as a very small quantity 
of the first order ; the second, being one-millionth of the first, must 
be regarded as a small quantity of the second order, and so on. Now, 
if we continue this series indefinitely, it is clear that we can make the 
terms become as small as we please without ever reaching absolute zero. 
It is also clear that, however small the terms of this series become, the 
ratio of any term to the one that immediately follows it is one million. 

10. Geometric Illustration of Infinitesimals. — The 

following geometric results will help to illustrate the theory 
of infinitesimals. 



GEOMETRIC ILLUSTRATION OF INFINITESIMALS. 11 



Let A and B be two points on the 
circumference of a circle ; draw the 
diameter AE, and draw EB produced 
to meet the tangent AD at D. Then, 
as the triangles EAB and ADB are 
similar, we have, 



and 



BE 


AB 


AE " 


" AD' 


AB 
AE = 


BD 
" AD' 



(1) 

00 




Now suppose the point B to approach the point A till it 
becomes infinitely near to it, then BE becomes ultimately 
equal to AE; but, from (1), when 

BE = AE, 
we have AB = AD. 

AB 

Also, -jv, becomes infinitely small, that is, AB becomes 

an infinitely small quantity in comparison with AE. Hence, 
from (2), BD becomes infinitely small in comparison with 
AD or AB ; that is, token AB is an infinitesimal of the first 
order, BD is an infinitesimal of the second order. 

Since DE — AE < BD, it follows that, token one side of 
a right-angled triangle is regarded as an infinitely small 
quantity of the first order, the difference between the kypotk- 
enuse and the remaining side is an infinitely small quantity 
of the second order. 

Draw BN perpendicular to AD ; then, since AB > AN, 

we have, 

AD - AB < AD - AN < DN ; 



therefore, 



AD-AB 
BD 



DN AD 
< BD < DE 



But AD is infinitely small in comparison with DE, there- 
fore AD — AB is infinitely small in comparison with BD : 



12 AXIOMS. 

but BD is an infinitesimal of the second order (see above), 
hence AD — AB is an infinitesimal of the third order. 

In like manner it may be shown that BD — BN is an 
infinitesimal of the fourth order, and so on. [The student 
who wishes further illustration is referred to Williamson's 
Dif. Cal,, p. 35, from which this was taken.] 

11. Axioms. — From the nature of an infinite quantity, 
a finite quantity can have no value when added to it, and 
must therefore be dropped. 

An infinitesimal can have no value when added to a finite 
quantity, and must therefore be dropped. 

If an infinite or an infinitesimal be multiplied or divided 
by a finite quantity, its order is not changed. 

If an expression involves the sum or difference of infinites 
of different orders, its value is equal to the infinite of the 
highest order, and all the others can have no value when 
added to it, and must be dropped. 

If an expression involves the sum or difference of infini- 
tesimals of different orders, its value is equal to the 
infinitesimal of the lowest order, and all the others can have 
no value when added to it, and must be dropped. 

These axioms are self-evident, and, therefore, axioms in the strict 
sense. For example, suppose we were to compare the mass of the sun 
with that of the earth : the latter weighs about six sextillion tons, the 
former weighs about 355000 times as much. If a weight of one grain 
were added to or subtracted from either, it would not affect the ratio 
appreciably ; and yet the grain, compared with either, is finite — it can 
be expressed in figures, though on the verge of an infinitesimal. If 
we divide this grain into a great many equal parts — a sextillion, for 
instance — and add one of these parts to the sun or the earth, the error 
of the ratio will be still less ; hence, when the subdivision is continued 
indefinitely, it is evident that we may obtain a fraction less than any 
assignable quantity, however small, which, when added to the sun or 
the earth, will affect the above ratio by a quantity less than any to 
which we can assign a value. 

By reason of the terms that may be omitted, in virtue of the prin- 
ciples contained in these axioms, the equations formed in the solution 



EXAMPLES. 13 

of a problem will be greatly simplified. It may be remarked that in 
the method of limits* when exclusively adopted, it is usual to retain 
infinitely small quantities of higher orders until the end of the calcu- 
lation, and then to neglect them on proceeding to the limit ; while, in 
the infinitesimal method, such quantities are neglected from the be- 
ginning, from the knowledge that they cannot affect the final result, 
as they necessarily disappear in the limit. The advantage derived 
from neglecting these quantities will be evident when it is remem- 
bered how much the difficulty in the solution of a problem is increased 
when it is necessary to introduce into its equations the second, third, 
and in general the higher powers of the quantities to be considered. 

EXAMPLES. 

1. Find the value of the fraction — - , if # is infinite, 

and a and b finite. 

Since a and b are finite, they have no value in comparison 

with x, and must therefore be dropped, giving us — = f 
as the required value of the fraction. 

2x a 

2. Find the value of the fraction ^r 7 , if x is infini- 
tesimal, and a and l finite. 

Since 2 is an infinitesimal, it has no value in comparison 
with a and h, and must therefore be dropped, giving us — -= 
for the required value of the fraction. 

8^-2 _i_ 2x 

3. Find the value of sn L » when x is infinite ; also 

AiX -4- X 

when x is infinitesimal. 

Ans. When x is infinite, 4; when infinitesimal, 2. 

a -n- i xi i - && + ^' 2 + cx + e i 

4. find the value of — = , when x is 

mx* -f- nx 2 -\- px -f q 

infinite; and when infinitesimal. 

Ans. When x is infinite, — ; when infinitesimal, -• 

in q 



* For a discussion of limits, see Chapter III. 



] 4 EXAMPLES. 

5. Find the value of -^ -. , when x is infinite; 

bx± — 4:X + 1 • 

and when infinitesimal. 

Ans. When x is infinite, go ; when infinitesimal, 2. 

a V A +1 1 ,4^ + 3^ + 2^-1 . 

6. Find the value of — — j-= - , when x is 

2^5 + 4:X 2 + 2x 

infinite; and when infinitesimal. 

Ans. When x is infinite, 0; when infinitesimal, go. 

7. Find the value of — t—= , when x is infinite ; and 

,.„.,., 4a 8 — mx 

when infinitesimal. 

Ans. When x is infinite, ; when infinitesimal, 7m. 

8. Find the value of - = , when x is infinite ; and 

2$ x 

when infinitesimal. 

Ans. When x is infinite, co ; when infinitesimal, 0. 

^ x 2?/ 

9. Find the value of j -" , when x and y are infini- 
tesimals. * ^ 

.4 as. We do not know, since the relation between x and y 
is unknown. 



CHAPTER II 

DIFFERENTIATION OF ALGEBRAIC AND TRANSCEN- 
DENTAL FUNCTIONS. 

12. Increments and Differentials. — If any variable, 
as x, be supposed to receive any change, such change is 
called an increment ; this increment of x is usually denoted 
by the notation Ax, read "difference x," or "delta x" where 
A is taken as an abbreviation of the word difference. If the 
variable is increasing, the increment is -+- ; but if it is 
decreasing, the increment is — . 

When the. increment, or difference, is supposed infinitely 
small, or an infinitesimal, it is called a differential, and is 
represented by dx, read "differential x," where d is taken as 
an abbreviation of the word differential, or infinitely small 
difference. The symbols A and d, when prefixed to a varia- 
ble or function, have not the effect of multiplication ; that 
is, dx is not d times x, and A# is not A times x, but their 
power is that of an operation performed on the quantity to 
which they are prefixed. 

If u be a function of x, and x becomes x+kx, the cor- 
responding value of u is represented by u-\-Aic ; that is, the 
increment of u corresponding to a finite increment of x is 
denoted by Aw, read " difference w." 

If x becomes x-\-dx, the corresponding value of u is rep- 
resented by u + dn; that is, the infinitely small increment 
of u caused by an infinitely small increment in x, on which 
u depends, is denoted by du, read " differential uP Hence, 
dx is the infinitesimal increment of x, or the infinitesimal 
quantity by which x is increased; and du is the correspond- 
ing infinitesimal increment of u. 



16 CONSECUTIVE POINTS — DIFFERENTIATION. 

The differential du or dx is. -f or — according as the 
variable is increasing or decreasing, i. e., the first value is 
always to be taken from the second. 

13. .Consecutive Values. — Consecutive values of a 
function or variable are values which differ from each other 
by less than any assignable quantity. 

Consecutive points are points nearer to each other than 
any assignable distance. 

Thus, if two points were one-millionth of an inch apart, they might 
be considered practically as consecutive points ; and yet we might have 
a million points between them, the distance between any two of which 
would be a millionth of a millionth of an inch ; and so we might have 
a million points between any two of these last points, and so on ; that 
is, however close two points might be to each other, we could still 
suppose any number of points between them. 

A differential has been defined as an infinitely small in- 
crement, or an infinitesimal; it may also be defined as the 
difference betiveen tivo consecutive values of a variable or 
function. The difference is always found by taking the first 
value from the second. 

In the Differential Calculus, we investigate the relations 
between the infinitesimal increments of variables from given 
relations between finite values of those variables. 

The operation of finding the differential of a function or 
a variable is called differentiation. 

14. Differentiation of the Algebraic Sum of a 
Number of Functions. 

Let u = v + y — z, (1) 

in which u, v, y 9 z, are functions of x.* 



* We might also, in a similar manner, find the differential of a function of sev- 
eral variables ; but we prefer to reserve the inquiry into the differentials of functions 
of several variables for a later chapter, and confine ourselves at present to functions 
of a single variable. 



DIFFERENTIATION OF A PRODUCT. ' 17 

Give to x the infinitesimal increment dx, and let du, dv, 
dy, dz, be the corresponding infinitesimal increments of u, 
v, y, z, due to the increment which x takes. Then (1) 
becomes 

u + du = v + dv -f y + dy — (z + <^0- (-) 

Subtracting (1) from (2), we have 

du = dv + <;/?/ — dz, (3) 

which is the differential required. 

Therefore, the differential of the algebraic sum of lk/\ 
any number of functions is found- by taking the alge- ' 
braie sum of their differentials. 

15. To Differentiate I — ^* 

y = ax±o. (1) 

Give to x the infinitesimal increment dx, and let dy be 
the corresponding infinitesimal increment of y due to the 
increment which x takes. Then (1) becomes 

y + dy = a (x + dx) ± b. (2) 

Subtracting (1) from (2), we get 

dy = adx, (3) 

which is the required differential. 



Hence, the differential of the product of a constant 
by a variable is equal to the constant multiplied by 
the differential of the variable ; cdso, if a constant be 
connected with a variable by the sign + or —, it dis- 
appears in differentiation. 

This may also be proved geometrically as follows : 

Let AB (Fig. 2) be the line whose equation is y = ax-\-b, 
and let (x, y) be any point P on this line. Give OM (= x) 



y 



18 



GEOMETRIC ILL USTRA TION. 



the infinitesimal increment MM' (=zdx), then the cor- 
responding increment of MP (=y) 
will be CP' (= dy). Now in the tri- 
angle CPP' we have 

CP = CPtanCPF;* 

or letting a = tan CPP', and substi- 
tuting for CP' and CP their values dy 
and dx, we have, 

dy = adx. 

It is evident that the constant b will disappear in differentiation, 
from the very nature of constants, which do not admit of increase, and 
therefore can take no increment. 

16. Differentiation of the Product of two Func- 
tions. 




Fig. 2 



Let 



u 



yz, 



(i) 



where y and z are both functions of x. Give x the infini- 
tesimal increment dx, and let die, dy, dz be the correspond- 
ing increments of u, y, and z. due to the increment which 
x takes. Then (1) becomes 

u -f- du = (y + dy) (z -f dz) 

±= yz + zdy + ydz -f dz dy. (2) 



Subtracting (1) from (2), and omitting dzdy, since it is 
an infinitesimal of the second order, and added to others of 
the first order (Art. 11), we have 

du = zdy -f ydz, frJ\ (3) 

which is the required differential. \ 

Hence, the differential of the product of two func- 
tions is equal to the first into the differential of the 
second, plus the second into the differential of the 
first. 



* In the Calculus as in the Analytic Geometry, the radius is always regarded as 
1, unless otherwise mentioned. 



GEOMETRIC ILLUSTRATIOX. 19 

This may also be proved geometrically as follows : 



Fig. 3. 



B a 



Let z and y represent the lines AB 
and BC respectively ; then will u rep- 
resent the area of the rectangle ABCD. 
Give AB and BC the infinitesimal in- 
crements Ba (='dz) and Cc (— dy) 
respectively. Then the rectangle ABCD 
will be increased by the rectangles ~BaCJt , 
DCbc, and Chcd, the values of which 
are ydz, zdy, and dzdy respectively ; therefore 

die = ydz + zdy + dz dy. 

But dzdy being an infinitesimal of the second order and 
connected with others of the first order, must be dropped 
(Art. 11) ; if this were not done, infinitesimals would not 
be what they are (Art. 8) ; the very fact of dropping the 
term dzdy implies that its value, as compared with that of 
ydz -\- zdy is infinitely small. 

The statement that ydz + zdy + dzdy is rigorously equal to ydz -zdy 
is not true, and yet by taking dz and dy sufficiently small, the error 
may be made as small as we please. 

.. Or, we may introduce the idea of motion, and consider 
that dz and dy represent the rate at which AB and BC are 
increasing at the instant they are equal to z aud y respec- 
tively. The rate at which the rectangle ABCD is enlarging 
at this instant depends upon the length of BC and the rate 
at which it is moving to the right -f- the length of DC and 
the rate at which it is moving upward. If we let dz repre- 
sent the rate at which BC is moving to the right, and dy the 
rate at which DC is moving upward at the instant that 
AB = z and BC = y, we shall have du = zdy + ydz as the 
rate at which the rectangle ABCD is enlarging at this in- 
stant. (See Price's Calculus, vol. i, p. 4-1.) ^V 



20 DIFFERENTIATION OF A FRACTION. 

17. Differentiation of the Product of any Num- 
ber of Fractions. 

Let u = vyz, (1) 

Then giving to x the infinitesimal increments, and letting 
du, dv, dy, dz be the corresponding increments of u, v, y, z, 
(1) becomes 

u + du = (v + dv) (y -f dy) (z -f dz). (2) 

Subtracting (1) from (2), and omitting infinitesimals of 
higher orders than the first, we have 

du = yzdv + vzdy + vy dz, i\f\ (3) 

and so on for any number of functions. >J 

Hence, the differential of the product of any num- 
ber of functions is equal to the sum of the products of 
the differential of each into the product of all the 
others. 

Con. — Dividing (3) by (1), we have 

du dv dy dz 

_ — h — H (4) 

u v y z w 

That is, if the differential of each function be di- 
vided by the function itself, the sum of the quotients 
will be equal to the differential of the product of the 
functions divided by the pi^oduct. 

18. Differentiation of a Fraction. 

x 
Let u = - , 

y 

then uy = x; (1) 

therefore, by Art. 16, we have 

udy f ydu = dx. 

Substituting for u its value, we have 



DIFFERENTIATION OF A POWER. 21 



- dy -f ydic = dx. 



\K 



Solving for du, we get 

ydx — xdy 

f ' '' 

which is the required differential. Al 

Hence, the differential of a fraction is equal to the 
denominator into the differential of the numerator, 
minus the numerator into the differential of the de- 
nominator, divided by the square of the denominator. 

Cor. 1. — If the numerator be constant, the first term in 
the differential vanishes, and we have 

xdy 



du = — 



f 



Hence, the differential of a fraction with a constant 
numerator is equal to minus the numerator into the 
differential of the denominator divided by the square 
of the denominator. 

Cor. 2. — If the denominator be constant, the second term 
vanishes, and We have 

7 dx 
du = — , 

y 

which is the same result we would get by applying the rule 
of Art. 15. 

19. Differentiation of any Power of a Single Va- 
riable. ^— 

Let y = *». * 

1st, Jllien n is a positive integer* I \ 
Eegarding x 11 as the product x, x, x, etc., of n equal fac- 
tors, each equal to x, and applying the rule for differentiating 
a product (Art. 17), we get 



22 DIFFERENTIAL OF A POWER OF A VARIABLE. 

dy = x n ~ l dx 4- x n ~ l dx -f x n ~ 1 dx + etc., to n terms. 

.*. dy = nx n ~ l dx. (1) 

2d. When it is a positive fraction. 

m 

Let y == X"; 

then y n — x m . 

Differentiating this as just shown, we have, 
ny n ~ l dy = mx m ~ l dx. 

Therefore, 



dy 


— 


m x m ~ 
n y n ~ ] 


i 
dx 




— 


m x m ~ 
n y h 






— 


m x m ~ 


m 

da 




n x 


m CU 


:. dy 


— 


m ™- 

— X* 

n 


dx. 



dx (since y n = x m ). 



3d. When n is a negative exponent, integral or 
fractional. 

Let y = x~ n ; 

then y = ■ — 

y x n 

Differentiating by Art. 18, Cor. 1, we have 

dy — ^ — = — nx~ n ~ l dx. (3) 

Combining the results in (1), (2), and (3), we have the 
following rule: The differential of any constant power 
of a variable is equal to the product of the exponent, 
the variable with its exponent diminished by unity, 
and the differential of the variable. 



EXAMPLES. 23 

Cor. — If n = \, we have from (1), 

dx 
dy = \x^~ x dx = \x~~*dx = — — • 

2Vz 
Hence, £7&e differential of the square root of a varia- 
ble is equal to the differential of the variable divided 
by twice the square root of the variable. 

examples. 

1. Differentiate y = 9 -f- 2x + x 3 -f- x 2 y B — x 7 -f 2axy. 

By Art. 14, we differentiate each term separately, and 
take the algebraic sum. By Art. 15, the constant 9 disap- 
pears in the differentiation ; and the differential of 2x is the 
constant 2, multiplied by the differential of the variable x, 
giving 2dx. By Art. 19, the differential of x 2 is 2x dx. The 
term x 2 y z is the product of two functions ; therefore, Art. 16, 
its differential is x 2 d(y s ) -\-y z d(x 2 ), which, Art, 19, gives 
ox 2 y 2 dy + 2y s xdx. In like manner proceed with the 
other terms, giving the proper rule in each case. The 
answer is 

dy = 2dx + 3x 2 dx + Zxhfdy + %xfdx — !x*dx 
+ 2ax dy + 2ay dx. 

2. u = ax B y 2 . du = Zax 2 y 2 dx + 2ax*y dy. 

3. u = 2ax - dx 2 — dbx* — 7. 

du = (2a — 6x — kaox 2 ) dx. 

4. u = x 2 y*. du = %x 2 y^ dy + 2xyi dx. 

5. u — 2lzr 2 + 3axhk 

7 - 2 i 7 3ax*dz 4:bdz 

du = dax*z* dx + — - — — — — =- • 

6. u = x* r . d u = -^i-. 



24 EXAMPLES. 



7. 


5 a , i 
u = ax* + — o — ox*. 

X 2 




du -{ iax '-^- w ~) dx - 


8. 


y 2 = 2px, to find the value of dy. 




dy = - dx. 

■ y 


9. 


a 2 y 2 + b 2 x 2 = a 2 b 2 , to find the value of dy. 




b 2 x 

dy = r- d#. 

a 2 y 


10. 


^ 2 + 2/ 2 = ^ to find the value of dy. 




dy = dx. 


11. 


a , ±ay dy 


U ~ l _ 2y2 aU ~ {b- %ff 


12. 


u = (a -1- bx + ca; 2 ) 5 . 



Regarding the quantity within the parenthesis as a varia- 
ble, we have, by Art. 19, 

du = 5(a +bx + cx 2 yd (a + bx + <%c 2 ) 
= 5 (« + bx + c# 2 ) 4 (& + 2cx) dx. 

13 w = 2x2 ~ 3 

4rX + # 2 

By Art. 18, 

__ (Ax + a 2 ) 6? (2a; 2 — 3) — (2a* — 3) d (4a: + x 2 ) 
U ~~ (4»-+ # 2 ) 2 . 

_ (jtaM-jc 2 ) 4a; da; — {t x 2 — 3) (4 + 2x) dx 
~ - {Ax + ik 8 ) 8 ■ 

(8a; 2 + 6a; + 12) <fo 



(4a + a; 2 ) 2 
2^ 4 _ SaW — Ax 5 _ 



EXAMPLES. 25 

1 -f a 7 (1 — 2x — x 2 ) clx 

15 - * = r+y ** = (i + V ~* 

16. %i — —- dy = — — r da. 

17. ?/ = (aa 2 — a 3 ) 4 . 

tfy = 4 (aa 2 — a 3 ) 3 (2aa — 3a; 2 ) dx. 

18. y == (a .+ fa«ji ^ = -^ (« + fo; 2 )f &*-& 



10 


# 7 6rta; 




* (P + a; 2 ) 3 ^ ~ (6 2 + a^ ** 


20. 


2/ = Vx* — a* (Art. 19, Cor.). 

(hl _ *(<*■—«*) _ 3a; 2 ^ 



21. y — W'Zax — a 2 , tfy — 



(a — x) dx 
\/2ax — x 2 



90 1 7 X & X 

22. y = 77r==- dl J = 



Vi - x 2 * (l _ ^)i 

23. y = V^+ v^W % = g + y 8ga? ^. 

2v# 

2 Vfl^ 4- foe + c 

25. y = (x 3 + a) (dx 2 + J) (Art. 16). 

% = (a 3 + a) d (3a 2 + b) + (3a 2 + J) rf (a 3 + ») 
= (15a* + 36a 2 + 6«a) tfa. 

26. y = (1 + 2a; 2 ) (1 4- 4a 3 ). 

dTjf = 4a (1 4- 3a 4- 10a 3 ) da. 

27. y = (a - a) V« 4- x. dy - - ~~==- 

2Va 4- x 

x / (a—3x)dx 

28. // = (a + &) \A - x. dt/ = > / — — - 

2 V a — a 



26 ILLUSTRATIVE EXAMPLES. 



ILLUSTRATIVE EXAMPLES. 

1. In the parabola y % = 4rr, which is increasing the most 
rapidly at x = 3, the abscissa or the ordinate ? How does 
the relative rate of change vary as we recede from the ver- 
tex ? 

2 
Differentiating y 2 =. 4#, we get dy = - dx, which shows 

that if we give to x the infinitely small increment dx, the 

2 
corresponding increment of y is - times as great ; that is, 

the ordinate changes - times as fast as the abscissa. At 
x = 3, we have y = V 12. Hence, at this point, 

dy = — -= dx = — ;= dx : 
Vl2 V3 

that is, the ordinate is increasing a little over one-half as 
fast as the abscissa at x — 3. 

At a? = l, y = 2, and dy = dx; that is, x and y are 
increasing equally; in general, at the focus the abscissa and 
ordinate of a parabola are increasing equally. At x = 4, 
y = 4:, and dy = \dx ; that is, «/ is increasing \ as fast as 
x. At x = 9, ?/ = 6, and ^ = -J^'; that is, ?/ is increas- 
ing J as fast as #. At # = 36, y = 12, and dy=z^dx; 
that is, ?/ is increasing i as fast as x. and so on. We see 

2 
from the equation dy = -dx, as well as from the figure of 

the parabola, that the larger x becomes, and therefore y, the 
less rapidly y increases, while x continues to increase uni- 
formly. 

2. If the side of an equilateral triangle is increasing uni- 
formly at the rate of J an inch per second, at what rate is 
its altitude increasing ? Is the relative rate of increase of 
the side and altitude constant or variable ? 



ILLUSTRATIVE EXAMPLES. 27 

Let x = a side of the triangle and y = its altitude. Then 

A/3 ' 
?/ 2 = \&, and % = — <^r, which shows that when x takes 

the infinitely small increment dx, the corresponding in ere- . 
ment of y is -y- times as great; that is, the altitude y 

always changes -— times as fast as the side x. When x is 
increasing at the rate of \ an inch per second, y is increas- 
ing — - times |, or -— inches per second. 

3. A boy is running on a horizontal plane directly towards 
the foot of a tower 60 feet in height. How much faster is 
he nearing the foot than the top of the tower ? How far is 
he from the foot of the tower when he is approaching it 
twice as fast as he is approaching the top ? When he is 
100 feet from the foot of the tower, how much faster is he 
approaching it than the top ? 

Let x = the boy's distance from the foot of the tower, 
and y = his distance from the top. Then we have 

y 2 = x 2 + 60 2 .' 

.-. dx = - dy ; 
x 

that is, the boy is nearing the foot - times as fast as lie is 
the top. 

2d. When he is approaching the foot of the tower twice 
as fast as he is the top, we have dx = 2dy, which in 

dx == - dy 

x J 

gives us y = 2x, and this in y 2 = x 2 + 60 2 gives us 

3x 2 = 60 2 , -• or x = -— — 34.64. 

A/3 



28 ILLUSTRATIV 

3d. When he is 100 ft 

y = y^OO i) = 116.62, 

, y 116.62 . . . . 7 . y , . 

and - = ^ nn ■ , which in ax = - #?/ gives 

.t 100 ^ ^ to 

<fe = 1.1662 6fe ; 

that is, he is approaching the foot of the tower 1.1662 times 
as fast as he is the top. 

4. In the parabola y 2 = Ylx, find the point at which the 
ordinate and abscissa are increasing equally; also the point 
at which the ordinate is increasing half as fast as the 
abscissa. Ans. The point (3, 6); and the point (12, 12). 

5. If the side of an equilateral triangle is increasing uni- 
formly at the rate of 2 inches per second, at what rate is the 
altitude increasing. Ans. V3 inches per second. 

6. If the side of an equilateral triangle is increasing uni- 
formly at the rate of 5 inches per second, at what rate is the 
area increasing when the side is 10 feet ? 

Ans. 15 a/3 sq. ft. per second. 

. V. . 

7. A vessel is sailing northwest at the uniform rate of 

10 miles per hour ; at what rate is she making north lati- 
tude? Ans. 7.07+ miles per hour. 

8. A boy is running on a horizontal plane directly toward 
the foot of a tower, at the rate of 5 mjles per hour • at what 
rate is he approaching the top of the tower when he is 60 
feet from the foot, the tower being 80 feet high ? 

Afls. 3 miles per hour. 



LOG A. EXPONENTIAL FUNCTIONS. 29 

LOGARITHM D EXPONENTIAL FUNC- 

TIONS. 

20. To Differentiate y — log x. — "We have 

/ dr 

y + dy = log (x + dx) = log x \1 + — 

= log x + log (l + -£). 

Subtracting, we have, 

, L dx\ /dx dx 2 \ 

dy = log \l+-J = m ^-— + etc.; 

(from Algebra, where m is the modulus of the system). 

dx 
.-. dy = d (log a?) = m — (Art. 11). 

27us result may also be obtained as follows: 

Let // = ax. (1) 

.*. log y = log a + log x. (2) 

By Art. 15, dy = a rfa\ (3) 

and J (log ?/) = <r/ (log a:). (4) 

Dividing (4) by (3), we get, 

d (log y) _ (7 (log x ) 
dy a dx 

= 18&4, from (1); 

or ^_G°i_ ? Zl = JL. 

d (log .r) " dx 

x 



30 DIFFERENTIAL OF A LOGARITHM. 

Multiply both terms of the second fraction by the arbi- 
trary factor m, and we have 

mdy 

* ( lQ g y ) = ~y~ (6) 

d (log x) " ' m dx' \- ' 

x 

"We may suppose m to have such a value as to make 



dQogy) = m-^-; 


(6) 


dx 
therefore, d (log x) = m — 

X 


(7) 


Similarly, let y = bz. 




:. log y — log h + log z. 


(8) 


Differentiating, dy = idz, 




and d (log y) == d (log z). 




Dividing and substituting, 




dy 
d (log y) _ y ^ 
d (log z) " dz 

z 




But d (log y) = m — 


(9) 


.*. 6? (log 2) = m — • 





In the same way we may show that the differential of the 
logarithm of any other quantity is equal to m times the 
differential of the quantity divided by the quantity, and 
hence the factor m is a constant, provided that the loga- 
rithms be taken in each case in the same system ; of course, 
if the logarithms in (8) be taken in a different system from 
those in (2), the numerical values of log y in the two eqna- 



DIFFERENTIA TI0N OF AN EXP ONENTIAL FUNCTION. 8 1 

tions are different, and therefore the m in (7) is different 
from the m in (9). Since m is a constant in the same sys- 
tem and different for different systems, it varies with the 
base of the system, as the only other quantities involved in 
logarithms are the number and its logarithm. That is, m is 
a function of the base; its value will be computed hereafter. 
(See Eice and Johnson's Calculus, p. 39 ; also, Olney's Cal- 
culus, p. 25.) 

Hence, the differential of the logarithm of a quan- 
tity is equal to the modulus of the system into the 
differential of the quantity divided by the quantity. 

Cor. — If the logarithm be taken in the Naperian * system, 
the modulus is unity, and we have 

, dx 

Hence, the differential of the logarithm of a quan- 
tity in the JVapeHan system is equal to the differen- 
tial of the quantity divided by the quantity. 



or 



21. To Differentiate y 


= a x . 


Passing to logarithms, we 


have, 


log?/ = 


x log a. 


Differentiating, we have 




dy 

y 


dx log a ; 


dy = 


y dx log a 
m 



a x 

dy = d (a x ) = — log a dx. 



* So called from the name of the inventor of logarithms ; also sometimes called 
natural logarithms, from being those which occur first in the investigation of a 
method of calculating logarithms. They are sometimes called hyperbolic logarithms, 
from having been originally derived from the hyperbola. 



32 DIFFERENTIATION OF AN EX' . CTION. 

Hence, the differential of Mai function 

with a constant base is equal c ; unction into the 

logarithm of the base into the a /erential of the ex- 
ponent, divided by the modulus. 

Cor. 1. — If we take Naperian logarithms, we have 
dy = d (a x ) = a x log a dx (since m = 1). 

Cor. 2. — If a = e, the Naperian base, then 

log a == log e = 1, 

and therefore dy = d(e x ) = e x dx. 

Sch. — In analytical investigations, the logarithms nsed 
are almost exclusively Naperian, the base of which system 
is represented by the letter e. Since the form of the differ- 
ential is the simplest in the l^aperian system, we shall in 
all cases understand our logarithms to be Naperian, unless 
otherwise stated. 

22. To Differentiate u = y*. 

Passing to logarithms, we have, 

log u = x log y. 
* 
Differentiating, we have, 

du , , dy 

— = log y dx + x — : 

u b * y 

or, du = u log y dx -f ux — ; 

.-. du = y x log y dx + xy x ~ l dy. 

Hence, to differentiate an exponential function 
ivith a variable base, differentiate first as though the 
base -tvere constant and the exponent variable, and 
second as though the base were variable and the expo- 
nent constant, and take the sum of the results. 



EXAMPLES. 



33 



EXAMPLES. 



^ 1. y = x log x. 

„2. y — log x 2 . 

^3. y — a losx . 

kA. y = a e \ 

f 5. y = x*\ dy = x^ 



dy = (log x + 1) f7a:. 
2^ 



% = 



« log * log ft dx 



dy = ft eT * * log o $& 

log x (1 + log a) + - x x dx. 
xj 



dy — 



xdx 



6. y = log v 1 — z 2 . 

^7. i/ = log (x + a/1 + «*)• 

v-^8. y = log (?-±-^) = log (ft + a) — log (ft — x). 

\ft — XI 



dx 



2adx 

^ ft 2 — f6' 2 



9. y = logA^\±^ = ilog(l + x)-ilog(l-x). 



dx 

dy — z s' 

^ 1 — x 2 



. 10. y = log (log x). 

11. ?/ = l°g' 2 #• 

- 12. y = ic*. 

13. y = log -— = 



% = 



efoj 



iC log X 
dx 



dy = 2 log a; 
dy =zz 5c« (log a -f 1) dx. 



^Jx 2 + 1 + X 

Multiplying both terms by the numerator to rationalize 
the denominator, we get 



34 LOGARITHMIC DIFFERENTIATION, 

y = log [a/^T"I — x]\ 
, 2dx 

14. y = e x (x — 1). dy = e x xdx. 

^15. y = £'(a: 2 — 2x + 2). ^ = eFtfdx. 

e x j 2p x 

17. y = e x log as. (??/ = e x flog a; + -) dx, 

18. 2/ = e '° g *^^\ 

Then log ?/ = log V« 2 4- a; 2 . 

/. 2/ = V« 2 + ^ 2 « 

-, JULLJl/ 



Va 2 + a; 2 



19. v = ^ * «V = t^ w 

v 1 + a; ^ (1 -f a;) 2 

_ A a; 7 dx 

+ 20. «/ = l°g _ • a# == 



Vs* + 1 + x * Vz 2 + 1 

c?a? 

21. y = log (s + a+VSaa + a?). dy = y ^~— ' 

22. y's=aj»V^. rfy = flV^^la^*^" 1 ^ 

23. Logarithmic Differentiation. — When the function 
to be differentiated consists of products and quotients, the 
differentiation is often performed with greater facility by 
first passing to logarithms. This process is called logarith- 
mic differentiation. 



EXAMPLES. 35 

EXAMPLES. 

1. u = x (a 2 + x 2 ) Va 2 — xK 
Passing to logarithms, we have 

log u = log x 4- log (a 2 + x 2 ) + \ log {a 2 — x 2 ). 

du dx 2xdx xdx 

= — + 



u ' ~ x a 2 -\- x 2 a 2 — x 2 
die = 



#2 /^2 i 2J 2 ) - 1 

ya 2 — a 2 J 
Va 2 



a 4 + oftc 8 — 4a 4 7 

or, aw = ■ — dx, 



1 4- x 2 

u = -v Passing to logarithms, we have 

j. — x 

log w = log (1 4- x 2 ) — log (1 — x 2 ). 

du "Zxdx 2xdx kxdx 



u 1 + ^ ' 1-^ (14- x 2 ) (1 — x 2 ) 
4:xdx 

3. u = (a* 4- I) 2 . du = 2a x (a x + 1) log a dx. 

a x — l , 2a aj logaaz 

t- aw = / - .x» ♦ 



4 



a* 4-1 (a x + l) 2 



a/1 4- » , dx 

du 



Vi — x (i — x) Vi — ^ 



ILLUSTRATIVE EXAMPLES. 

1. Which increases the more rapidly, a number or its 
logarithm? How much more rapidly is the number 4238 
increasing than its common logarithm, supposing the two 
to be increasing uniformly ? While the number increases 
by 1, how much will its logarithm increase, supposing the 



36 ILLUSTRATIVE EXAMPLES. 

latter to increase uniformly (which it does not) while the 
number increases uniformly. 
Let x = the number, and y its logarithm ; then we have 

y == logs; 

.*. cli/ = — dx, 

J x ' 

which shows that if we give to the number (x) the infinitely 
small increment (dx), the corresponding increment of y is 

ffl 

— times as great; that is, the logarithm (y) is increasing 

x 

771 

— times as fast as the number. Hence, the increase in the 
x 

common logarithm of a number is '•>, =, < the increase 
of the number, according as the number (x) <, ==, > the 
modulus (m). 

When x = 4238, we have 

m __ .434294 48 

c ^ x — Toob dx; 



* 4238 4238 

4238 
hence, dx — AnAnnAAC ; dy = about 9*758 dy; 
9 .43429448 ^ * ' 

.43429448 
that is, the increment of the logarithm is '- — j— l — part of 

the increment of the number, and the number is increasing 
about 9758 times as fast as its logarithm. 

While the number increases by 1, its logarithm will in- 
crease (supposing it to increase uniformly with the number) 

- — — — — times 1 = .00010247; that is, the logarithm of 
4238 

4239 would be .00010247 larger than the logarithm of 4238, 

if it were increasing uniformly, while the number increased 

from 4238 to v 4239. 

Remark. — While a number is increasing uniformly, its logarithm 
is increasing more and more slowly ; this is evident from the equation 

dy = — dx, which shows that if the number receives a very small in- 
x 



TRIGONOMETRIC FUNCTIONS, 37 

crement, its logarithm receives a very small increment ; but on giving 
to the number a second very small increment equal to the first, the 
corresponding increment of the logarithm is a little lew than the first, 
and so on ; and yet the supposition that the relative rate of change of 
a number and its logarithm is constant for comparatively small changes 
in the number is sufficiently accurate for practical purposes, and is the 
assumption made in using the tabular difference in the tables of loga- 
rithms. 

2. The common logarithm of 32T is 2.514548. What is 
the logarithm of 327.12, supposing the relative rate of 
change of the number and its logarithm to continue uni- 
formly the same from 327 to 327.12 that it is at 327 ? 

Ans. 2.514707. 

3. Find what should he the tabular difference in the table 
of logarithms for numbers between 4825 and 482G ; in other 
words, find the increment of the logarithm while the num- 
ber increases from 4825 to 4826. Ans, .0000900. 

4. Find what should be the tabular difference in the table 
of logarithms for numbers between 9651 and 9652. 

Am, .0000450. 

5. Find what should be the tabular difference in the table 
of logarithms for numbers between 7235 and 7236. 

Ans, .0000601. 



TRIGONOMETRIC FUNCTIONS. 
24. To Differentiate y = sin x. (1) 

(live to x the infinitely small increment dx. and let dy 
represent the corresponding increment of y ; then we have 

y -\- dy = sin (x -f dx) 

= sin x cos dx -f cos x sin dx. (2) 

Because the arc dx is infinitely small, its sine is equal to 
the arc itself and its cosine equals 1 ; therefore (2) may be 
written 

y -f- dy = sin x + cos x dx. (3) 



38 TRIGONOMETRIC FUNCTIONS. 

Subtracting (I) from (3), we have 

dy = cos x clx. (4) 

Hence, the differential of the sine of an are is equal 
to the cosine of the arc into the differential of the arc. 

25. To Differentiate y = cos x. 

Give to x the infinitely small increment dx, and we have 
y -f dy = cos (x + dx) 

= cos x cos dx— sin x sin dx 
= cos x — sin x dx (Art. 24). 
.*. dy == — sin a; da\ 

Otherwise thus: 

We have y == cos x = sin (90° — x). 

Differentiating by Art. 24, we have 

dy = cos (90° — x) d (90° — x) 
= sin as rf (90° — a). 
.♦. dy — — sin * tfo. 

Hence, £foe differential of the cosine of an arc is 
negative and equal to the sine of the arc into the dif- 
ferential of the arc. (The negative sign shows that the 
cosine decreases as the arc increases.) 

26. To Differentiate y = tan x. 

Tir i sin z 

We have y = tan x = 

gby A 
dy = 



COS X 

Differentiating by Arts. 18, 24, and 25, we have 

cos x d sin x — sin x d cos x 
cos 2 x 



cos 2 x -f sin 2 x , $» 

-= rfiC = s— 

cos^ a; cos" 1 a; 

sec 2 x dx. :. dy = sec 2 x dx. 



TRIGONOMETRIC FUNCTIONS, 39 

Otherwise thus: 

Give to x the infinitesimal increment dx, and we have 

y 4- dy = tan (x + dr) 

.*. dy = tan (.t -f- efc) — tan x 
tan a; -f tan cfo 



1 — tan x tan «te 



— tan x 



tan a; -f dx , . . _ _ . 

= - t , tan re (since tan dx = dx) 

1 — taxi x ax ' 

dx + tan 2 x dx 

= — : t— = sec 2 x dx 

1 — tan x dx 

(since tan x dx, being an infinitesimal, may be dropped from 
the denominator), 

.-. dy = sec 2 x dx. 

Hence, the differential of the tangent of an are is 
equal to the square of the secant of the arc into the 
differential of the arc. 

27. To Differentiate y = cot x. 

We have y = cot x = tan (90° — x). 

.: dy = sec 2 (90° — x) d (90° — x). 

,\ dy = — cosec 2 x dx. 

The minus sign shows that the cotangent decreases as the arc 
increases. 

Hence, the differential of the cotangent of an arc is 
negative, and equal to the square of the cosecant of 
the arc into the differential of the arc. 

28. To Differentiate y = sec x. 

1 



We have y = sec x z= 



cos x 



40 TRIGONOMETRIC FUNCTIONS. 

cl cos x sin x rfx 

.-. ay = = — = 7 y — = sec x tan x ax. 

u cos** x cos^ x 

Hence, the differential of the secant of an are is 
equal to the secant of the same arc, into the tangent 
of the arc, into the differential of the arc. 

29. To Differentiate y = cosec x. 

We have y = cosec x = sec (90° — x), 

.'. dy = cl sec" (90° — x) 

= sec (90° — x) tan (90° — x) cl (90° — x) 
= — cosec x cot x clx. 

Hence, the differential of the cosecant of an arc is 
negative, and, equal to the cosecant of the arc, into the 
cotangent of the are, into the differential of the arc. 

30. To Differentiate y = vers x. 

We have y = vers x = 1 — cos x. 

.-. dy = cl (1 — cos x) = sin x clx. 

Hence, the differential of the versed-sine of an are 
is equal to the sine of the are into the differential of 
the arc. 

31. To Differentiate y = covers x. 

We have y = covers x = vers (90° — x), 

.-. dy == cl vers (90° — x) = sin (90° — x) d (90° — x) 
— — cos x clx. 

Hence, the differential of the cover sed-sine of an 
are is negative, and equal to the cosine of the arc into 
the differential of the arc. 



GEOMETRIC • DEMONSTRA Tl ON. 



41 



32. Geometric Demonstration. — The results arrived 
at in the preceding Articles 
admit also of easy demonstra- 
tion by geometric construction. 

Let P and Q be two consec- 
utive points* in the arc of a 
circle described with radius == 1. 
Let x = arc AP ; then 




dx = arc PQ. 
From the figure we have, 


O N 

Fig. 4. 


PM =± sin x; 


XQ = sin (x -j- dx) ; 


.-. QR 


= d sin x. 


OM = cos x; 


ON = cos (x + dx) ; 



.*. NM = — d cos x (minus because decreasing). 

AT = tan x ; AT 7 = tan (x + dx) ; 

.-. TT' = d tan x. 

OT = sec x) OT' = sec (x + dx) ; 

.-. DT' = d sec x. 

Now, since RP and QP are perpendicular respectively to 
MP and OP, and since DT and TT' are also perpendicular 
to OT and OA respectively, the two infinitely small triangles 
PQR and DTT' are similar to MOP. Hence we have the 
following equations : 

d sin x = RQ = QP cos PQR 
= cos x dx. 

.'. d sin x = cos x dx. 



* All that is meaut here is that P and Q are to he reasoned upon as though they 
were consecutive point? ; of course, strictly speaking, consecutive points can never 
be represented geometrically, since their distance apart is less than any assignable 
distance. When Ave say that P and Q are consecutive points, we may regard the 
distance PQ in the figure as representing the infinitesimal distance between two 
consecutive points, highly magnified. 



42 GEOMETRIC DEMONSTRATION. 

d cos x = — PE = — PQ sin PQR 
= — sin sb «fe. 

.\ 6? cos x = — sin x dx. 

d tan a; = TT' = DT sec DTT' = DT sec x 
= OT-QP sec a; (since DT = OT-QP) 
= sec 2 x dx. 

.'. (? tan a? = sec 2 a; eto. 

rf sec a; = DT' = DT tan DTT' 

= OT-QP tan x = sec x tana; dx. 

.'. d sec x = sec a; tan x dx. 

Also, c5 = — d (cot a?), 

and cd = — J (cosec x). 

But the triangle cbd is similar to the triangle OPM, since 
cb and db are respectively perpendicular to MP and OP. 
Hence we have 

d cot x = — cb = — db cosec dc# 

= - - bO • QP cosec x = — cosec 2 a? $& 

.*. d cot a? == — cosec 2 a 1 rfa. 

d cosec a; = —cd== — f?# cot rfci 
= — OJ-QPcota; 
= — cosec x cot x dx. 

,\ d cosec x = — cosec a; cot a; £?a?. 

From the figure we see that the differential of the versed- 
sine is the same numerically as that of the cosine, but with 
a contrary sign, i. e., as the versed-sine increases the cosine 
decreases ; also the coversed-sine has the same value nu- 
merically as the sine. (See Price's Calculus, Vol. I, p. 64.) 



EXAMPLES. 43 

EXAMPLES. 

^ \. y = sin mx. By Art. 24 we have, 

dy = cos mx d {mx) = m cos mx dx. 

- 2. y = sin (a; 2 ). 

<&/ = cos (a 2 ) d(a 2 ) = 2a: cos (a: 2 ) Jrc. 

"-3. y = sin w a. 

v dy = m sin m_1 a: r? (sin $) = m sin" 1-1 a cos jc dx. 

4. «/ = cos 3 x. 

dy = 3 cos 2 a: d (cos a) = — 3 cos 2 a sin x dx 
== 3 (sin 3 a — sin a:) tfz. 

w 5. y = sin 2a: cos x. 

dy = sin 2a; d cos a; + cos x d sin 2a: 

= — sin 2x sin a; dx + 2 cos 2a: cos x dx. 

6. ?/ = cot 2 (a 3 ). dy = — 6a 2 cot a; 3 cosec 2 a^ da:. 

7. y = sin 3 a cos x. dy = sin 2 a- (3 — 4 sin 2 x) dx. 

8. y = 3 sin 4 a:. d# = 12 sin 3 a: cos a* dx. 
v^ 9. y = sec 2 5a. dy = 10 sec 2 5a- tan 5a: dx. 

10. y = log sin x. 

_ d (sin x\ . . , _ x cos a: 7 _ 

r/y = -A (Art. 20) = -r— dx = cot a: da:. 

* sm a; v 7 sin a: 

11. ^/ = log (sin 2 x) =3 2 log sin a. dy = 2 cot a: da:. 

12. i/ — l°g cos ^ dy = — tan a cZa:. 

6? tan a: 2dx 



13. 


y = log tan a\ 


J " tan a: " sin 2a: 


14. 


y = log cot a\ 


2da 
J sin 2a: 


15. 


y = log sec a;. 


dy = tan x dx. 


16. 


?/ = log cosec a. 


dy = — cot a: da:. 



44 ILLUSTRATIVE EXAMPLES. 

17. 



= l0 Vr 



+ COS X 



COS X 

= -J- log (1 + cos x) — \ log (1 — cos a). 



7 dx 

dy = 



sm a; 

18. y = e x cos x. dy = e x d cos # -f cos x de x 

= — e x sin x dx + e x cos a; <£z 
= e x (cos a; — sin #) r/a:. 

19. y = x sin # -f cos a;. dy = x cos a; *?#. 

20. ?/ = xe cosx . dy — e C3SX (1 — z sin a) dx. . 

21. 2/ — eC0SX sni - T - <% — eCOSX ( cos # — sm2 x ) d x * 
_/ 22. ?/ = log Vsin a* + log a/cos a; 

= J log sin x + J log cos #. 

.-. ^ = ^ (cot a; — tan x) dx = cot 2x dx. 

k^ 23. y = log (cos a: + V— 1 sin a;), t/y == V — 1 ^#. 



, 1 + sin x dx 

24. y = log \ / = ~t—-- tfy = 



25. ?/ = lo g tan ( 45 ° + i x ) 



— sm a; cos x 

dx 



cos a: 



26. ?/ = sin (log x). dy = - cos (log x) dx. 



ILLUSTRATIVE EXAMPLES. 

1. Which increases faster, the arc or its tangent? When 
is this difference least, and when greatest? What is the 
value of the arc when the tangent is increasing twice as 
fast as the arc, and when increasing four times as fast as the 
arc ? 

From y = tan x, we get dy = sec 2 x dx, which shows 
that if we give to the arc (x) the infinitesimal increment dx, 



ILLUSTRATIVE EXAMPLES. 45 

the corresponding increment of the tangent (y) is seed- 
times as great ; that is, the tangent (y) is increasing secant 
square times as fast as the arc, and hence is generally in- 
creasing more rapidly than the arc. When x = 0, sec x == 1 ; 
therefore, at this point, the tangent and the arc are increas- 
ing at the same rate. When x = 90°, the secant is infinite ; 
therefore, at this point, the tangent is increasing infinitely 
faster than the arc. 

When the tangent is increasing twice as fast as the arc, 
we have dy = 2dz 9 or sec 2 # = 2, which gives x = 45° ; 
hence at 45° the tangent is increasing twice as fast as the 
arc. 

When the tangent is increasing four times as fast as the 
arc, we have dy = \dx, or sec 2 x = 4, which gives x = 60° ; 
hence at 60° the tangent is increasing four times as fast as 
the arc. 

2. Assuming that the relative rate of increase of the sine, 
as compared with the arc, remains constantly the same as 
at 60°, how much does the sine increase when the arc in- 
creases from 60° to 60° 20'. 

Let x = the arc and y its sine ; then we have y = sin .r, 
.-. dy = cos xdx, which shows that the increment of the 
sine is cosine times the increment of the arc. Now the arc 

of 20' = fl* 1 ^ = -0058176 = dx; therefore, 

loU X o 

dy = cos 60° dx = % x .0058176 = .0029088, 

which is the increase of the sine on the ahove supposition, 
and is a little greater than the increase as found from a table 
of natural sines, as it should be. since we have supposed the 
sine to increase uniformly while the arc was increasing 
uniformly from 60° to 60° 20', whereas the sine is increasing 
more and more slowly while the arc is increasing uniformly. 
This is evident from the equation dy =. cos x dx, and also 
from geometric considerations. ' 



46 CIRCULAR FUNCTIONS. 

3. The natural cosine of 5° 31' is .995368. Assuming 
that the relative rate of change of the cosine and the arc 
remains the same as at this point, while the arc increases to 
5° 32', what is the cosine of 5° 32' ? Ans. .995340. 

4. The logarithmic sine of 13° 49' is 9.3780633. Assum- 
ing that the relative rate of change of the logarithmic sine 
and the arc remains the same as at this point, while the arc 
increases to 13° 49' 10", what is the logarithmic sine of 
13° 49' 10"? Ans. 9.3782060. 

5. The log cot 58° 21/ = 9.789863. On the same sup- 
position as above, what is the decrease of this logarithm for 
1 second increase of arc. Ans. .00000471. 

6. A wheel is revolving in a vertical plane about a fixed 
centre. At what rate, as compared with its angular velocity, 
is a point in its circumference ascending, when it is 60° 
above the horizontal plane through the centre of motion. 

Ans. Half as fast. 

CIRCULAR FUNCTIONS. 

33. To Differentiate y — sin -1 x.* 

We have x = sin y ; 

therefore, dx = cos y dy = V(l — sin 2 y) dy 

== Vl — # 2 dy. 
dx 

•'• d v = vT=^ = d ( sin-1 ^ 

34. To Differentiate y — cos -1 x. 
We have, x = cos y ; 



therefore, dx = — sin y dy = — yl — cos 2 ?/ dy 

= — a/1 — x 2 dy. 
.-. dy = ~ — — = d (cos -1 x). 

* This notation, as already explained, means y = the arc whose sine is x. 



CIRCULAR FUNCTIONS. 47 

35. To Differentiate y = tan -1 x. 

We have x =± tan y ; > 

therefore, dx = sec 2 y dy == (1 + tan 2 y) dy 

••■ ^ = r^ = fZ ( tan_1 ^ \ 

36. To Differentiate y = cot -1 x. 

"We have $c = cot y ; 

therefore, dx = — cosec 2 y dy = — (1 + cot 2 ?/) dy 

= - (1 + x 2 ) dy. 



f^A 



dx 

1 +~^ 



37. To Differentiate 2/ = sec 1 a?. 

We have a? = sec y ; 

therefore, dx = sec y tan y dy = sec y Vsec 2 y — 1 dy 

= x\/x 2 — 1 dy. 

38. To Differentiate y = cosec -1 x. 

We have x = cosec y ; 

therefore, . dx = — cosec ?/ cot y d?/ 



= — cosec y V cosec 2 y — 1 dy 



= — zv^ 2 — 1 dy. 
dx 



48 EXAMPLES. 

39. To Differentiate y = vers -1 x. 

We have x = vers y ; 

therefore, dx = sin y dy = a/1 — cos 2 2/ % 



= V 1 — (1 — vers y) 2 dy 
= a/2 vers y — vers 2 y dy 



= <\/2% — x z dy. 

,\ dy = — = d (vers -1 x). 

V%% — x 2 

40. To Differentiate y = covers 1 x. 

"We have x = covers y ; 

therefore, dx = —cosy dy = — V± — sin 2 y dy 
= — a/1 — (1 — covers yf dy 



(A 



= — v 2 covers ?/ — covers 2 ?/ ^ 

= — ^/%x — x 2 dy. 

,\ dy = - — — ■ = rZ (covers" 1 x), 

V'Zx — cc 2 

EXAMPLES. 

1. Differentiate y = sin -1 -• 

We have, by Art. 33, 

x dx 

1 a a dx 

dy = 



H 



V' -I 



x 2 \^a 2 — x 2 
a 2 



x 

2. Differentiate y r = a sin -1 -« 

. a 



EXAMPLES. 

dx 

a — 

n 

We have, Art. 33, dy — 



49 



adx 



v 7 -? 



V a 2 — x* 



Geometric illustration of Examples 1 and 2 : 

Let OA = 1, OA' = a, y = arc AB, y' = arc A'B!, 
x = M'B'. 

B'M' 



Now 




BM 


= 


OB' 


a 


9 




arc 


AB 


= 


sin -1 
sin -1 

sin" 1 


BM 

B'M' 

OB' 

X 

a 










= 


y (see Ex. 


1) 




Fig. 5. 



M A M' K 



A'B' = A'O • arc AB = A'O . sin" 1 



B ; M' 
OB' 



= a snv 



■i - = y (see Ex. 2). 



Also, A'B' — sin -1 B'M' = sin -1 x (to radius a) 

x 
.'. a sin -1 - (to radius 1) = sin -1 x (to radius a). 

Hence, in Example 1, y is the arc AB (to radius 1), and 

x 
is given in terms of the sine - (to radius 1) ; while in 

Example 2, y is the arc A'B' (to radius a), and is given in 

% 
terms of the sin - (to radius 1). 
a v ' 

If we give B'M' (which is x in both examples) an incre- 
ment (= dx), the corresponding increment in A'B' will be 
a times as great as that on AB ; that is, dy' in Ex. 2 is a 
times dy in Ex. 1. 
3 



50 



\J 3. y = cos -1 — 

J 4. y = tan -1 — 
* a 



5. v = cot -1 — 



J 6. y = sec -1 — 
9 a 

' 7. y = cosec -1 -' 



8. y = vers -1 — 
* a 



x 
9. y = covers -1 * 



iujruM/0. 




dy = 


dx 


Va 2 — x 2 


dy = 


adx 
a 2 + x 2 


rfy == 


adx 
a 2 -f x 2 


dy = 


adx 


X^/ X 2 _ a 2 


dy = 


adx 


x Vx 2 — a 2 


dy = 


dx 




V%ax — x 2 


dy = 


dx 



a V%ax 



i j 10. y = a cos -1 -• 

adx 
, a adx 



s/ 



x 2 "s/a 2 



dx 

■x n a a 2 dx 

11. y = a tan -1 — dy = 



a " x 2 a 2 + x 2 

+ a 2 



dx 
a 

12. y — a cot -1 -• dy = 



x a a 2 dx 



a J x 2 a 2 + x 2 

+ a 2 

dx 

< x n a a 2 dx 

13. y — a sec -1 -■ J" 



x /x 2 
a\ a 2 



a T 7^2 ' X A/ rgi 



MISCELLANEOUS EXAMPLES. 51 



X 

14. y = a cosec -1 -• 
* a 



dx 

7 # a 2 dx 

dy - — 



x lx 
ay a 

or. 

16. y = a covers -1 



2 %<\/z 2 — a 2 

2 ~~ *- 



t x , a adx 

15. y = a vers -1 - • dy 



x x 2 V%ax — x 2 

a a 2 
x 
a 



dx 

■j a adx 

dy = 



\M 



V2ax — x 2 



MISCELLANEOUS EXAMPLES. 



1. 


y ■ 
y ■ 

y 
y 

y 


a + x 
Va — x 


dy = 

X 2 . 


3« — x ., 

: -,dx. 

2 (a — xj* 




/ 




2. 


= j/ x — Va 2 — 
dy : 

X 






(x + \/a 2 — £ 2 )f/# 


3. 

4. 


dy = 
dy = 


dx 


x + y'l _ ^2 

a* 


' 2x (1 — x 2 ) + ^\ZTx 2 

3x 2 dx 
V " (i _ ^2)t 

-a* (a 2 — 2x 2 )dx 


5 


V(i - ^) 3 




" sy^ - — ~& 


2x 2 {a 2 - z 2 )t 



52 MISCELLANEOUS EXAMPLES. 

Vx 2 + 1 — x 



6. y = 



Vx 2 + 1 + x 



In fractions of this form, the student will find it an ad- 
vantage to rationalize the denominator, by multiplying both 
terms of the fraction by the complementary surd form ; that 
is, in the present case, by Vx 2 + 1 — x. Thus, 



Vx 2 + 1 — x vx 2 4- 1 — - x 
II = — x — - 

Vx 2 + 1 + x Vx 2 + 1 — X 

_ (V^~+T — xf 
_ _ 

.-. dy = 2 (Vx*~+T - x) (--JL= — l\ dx 

\Vx 2 + 1 / 

= 2(2x-^t±£\dx. 
V Vx* + 1/ 






Vl — Vx _ Vl— a; 

4- V* Vi + V^ X + ^ 



dy = — 



2 (l + V») V# — # 2 



\A 4- a? -f a/1 — # 

a/i +%— Vi — # 



dy = J . — dx. 



== Vx • y Vx + 1. 

12V x 2 ' \/ Vx + 1 



7V^ + 4 , 

dy = —=zdx. 



MISCELLANEOUS EXAMPLES, 



53 



10. y = \J 



a--~ + V{o 2 - x*y 
\x 



3b 



4rX 



dy 



2xVx V(c 2 — x 2 ) 



dx. 



4\/fl- -~ + V(c*-x*)' 
V vx 

11. y = \J 2x—\—\/2x—\ — ^/2x—\ — etc., ad inf. 
Squaring, we have, 



— 2x— 1 — y 2x—l—y2x—l — V2x—l— etc., ad inf 
if — 2x — 1 — y ; 
y = —i± W 8x — 3. 
.*. rfy = ± 



^/ 4 

Hence, 
and 



2" -n 2" 
2ffa 



V8.T - 3 



12. y 



yl + a? + VI — a: 2 
vT+~a? — \/l— 2? 



^ 



~ ^( 1 + vr=^) 



rfo. 



13. y = log — 



<fy = 



dx 



, 1 . . \/l + * + Vl — « 

^ 14. ?/ = log - ~75=- <fy = " 



x (1 + a: 2 ) 
dx 



yl + £ — Vl — a5 



15. 



log 



V # 2 + a 3 — a: 



7 ^ , 



^V'l-x 2 
dx 



x \/x 2 + a? 



54 MISCELLANEOUS EXAMPLES. 

16. y = log [a/F+^2 + yT^T 2 ]. 

-y iXJU (A/JO 

X Z\/l — x* 

17. y^log^-a)- ^-^ . ^ = |±| f &. 

18. y = a x \ dy = 2a x * log a xdx. 

19. y = e x (1 — a; 3 ), dy = e x (1 — 3x 2 — x*)dx. 

e x — e~ x 7 Adx 

20. ?/ = — -• dy 

v oX _i_ />— iC «? 






gar g— a; 

21. y = log (e* + er*). dy = dx. 



1 7 af (1 ~ log a;) , 

22. y = x° . d*/ = — * g-4-Zdk 

23. # = 2e ^(al - 3a? + 6a* - 6). ^ = a?e ^'dfc 

24- y = 7- ( !X 1)f ,f (See Art. 23.) 
(a; — 2)4 (a; — 3)« 

(a- _ 1)1 (7a; 2 + 30a; - 97) , 

dy — — ± — = i-^-dx- 

12 (x - 2)* (x - 3)V- 

25 . y = (-±^J^3l. (Art. 23.) 
3 (x + 2) 4 v 

% == 



x Vx'-l 

36 - * = *— ys^r*" 

27. y = sin a; — -J- sin 3 a:. 

28. y = i *^ an3 ^ — tan a; + a;. 

29. y = i tan 3 a; + tan x. 



X 2 (x + 3)1 


dx 




(s + 2)« (a; + 1)* 




2(Va 2 — 1)4 


v yx, — j 


[ dx 


x+ Vx* 


^T 




dy = 


cos 3 a 


dx. 


dy = 


tan 4 a 


j (7a?. 


dy = 


sec 4 a 


rfa?. 



MISCELLANEOUS EXAMPLES. 55 

30. y = sin e x . dy = e x cose* dx. 

31. y = tan 2 x -f log (cos 2 a;), dy = 2 tan 3 a; dx. 

32. y = log (tan a; -f sec x). dy = sec x dx. 

sin a; ' (cos 3 # — sin 3 x) . 

33. v = - dy = ^-. ~dx. 

J 1 + tan x * (sin x + cos xf 



. , /a cos x — b sin a; 

34. </ = io g y- 



35. 


y = tan e*. dy 


36. 


y — tan Vl — x. dy 


37. 


y = x™\ dy = x i[a * ( 


38. 


2 3 cos x 


9 ' sin 2 x cos x sin 2 x 


39. 


y = sin x — 



cos # 4- & sin x 

j — rib dx 

" a 2 cos 2 x — W- sin 2 x 

L L 

e x sec 2 e x dx 

(sec V I — a:) dx 
2V1 — x 

, sin j»\ - 
c • losr a; H da;. 



3 log tan |. 

sin 3 a; cos 2 a; 



Wehave, rfy = d (^=) - ^A " I " 



-fa; 2 
cfcr 1 dx 



(1 + ^2)1 ' (i + ^)i 1 + a? 2 
\( 40. y = cos -1 x Vl — x 2 = cos -1 Vx 2 — x*. 



We have, dy = — d Vx 2 — x 4 -r- y 1 — (a: 2 — a; 4 ) 
(1 — 2a; 2 ) dx 



Vi 



+ Vl — x* 4- 3*. 



56 MISCELLANEOUS EXAMPLES. 

(1 — 2z 2 ) dx 



/. dy 



V (1 — £ 2 ) (1 — * 3 + 3*) 

2<£z 



/ 41. 2/^sin-K^A/l-^). 

3<i# 
^ 42. # == sin -1 (Sx — 4a; 3 ). dy = — 

/ V 1 — # 2 

■n.44 ^sm-^—. #= ___. 

45. a; = vers -1 «/ — V%ry — # 2 (where vers -1 y is taken 
to radius r). 

We may write this (see Art. 40, Exs. 1 and 2), 

x = r vers -1 - — V%ry — y % . 

.. _ rdy rdy — ydy 

(XX — ;' ' — — ■ ~~ , 

V2ry — y 2 V%ry — y % 

ydy 

V%ry — y 2 

46. y = Va? — a? -f « 2 sin" 1 -• % = 2\/a 2 — x\ 

47. y = log y -j-^l + i tan -1 a. 

48. v =r vers -1 — • 
9 9 

49. y = e tan_,a . 



(• 


dy = i 


dx 


dy 


dx 




" V$x~^ 


-~7 2 


dy 


~ 6 1 


dx 

~+~x* 



MIS CEL LA XE US EXAMPLES. 



57 



50. y = x slD ' 1 x . 

dy = af 1 ' 

-51. y = sec -1 nx, 
-52. y = sin -1 



"# log a; + (1 — a; 2 ) '* sin" 1 a; ' 

_ _ X (1 - a; 2 )* 

dx 

x y / 7i z x 2 — 1 



\dx. 



Va 2 + 



dy = 



adx 
a 2 + x 2 ' 



53. y = sin -1 A/sin a:. dy = % (a/1 + cosec a:) <££. 

„ . /l — cos a; , , . 

54. y = tan-y i - rcosV ^ = ffe. 



55. ?/ = 



x sm _1 a; 



-f- log v 1 — a: 2 , c??/ = - 



sin -1 a; 



Vl — x 2 (1— a; 2 )t 

56. ?/ — ( x + a ) tan -1 A / Vctx. 

dy = tan" 1 a A 



dx. 



dx. 



57. 2/ = sec_1 1 



V& 



XV D 



58. 2/ 



tan" 



^V^ 2 + x 
Sa 2 x — X s 



a z — Sax 2 



dy = 



dy 



dx 



x^x> + x — l 
oadx 



a 2 + x 2 



59. y = sin 



— sin -1 . 



Va(l + x 2 ) 

dy == 



V« 



(1 + ;t 2 ) y^ + fo8 



efo. 



60. If two bodies start together from the extremity of the 
diameter of a circle, the one moving uniformly along the 
diameter at the rate of 10 feet per second, and the other in 
the circumference with a variable velocity so as to keep it 
always directly above the former, what is its velocity in the 



58 MISCELLANEOUS EXAMPLES. 

circumference when passing the sixtieth degree from the 

starting-point? . 20 „ , , 

ft r Ans. — - feet per second. 

61. If two bodies start together from the extremity of the 
diameter of a circle, the one moving uniformly along the 
tangent at the rate of 10 feet per second, and the other in 
the circumference with a variable velocity, so as to be always 
in the right line joining the first body with the centre of 
the circle, what is its velocity when passing the forty-fifth 
degree from the starting-point. Ans s 5 ft. per second. 



CHAPTER III 

LIMITS AND DERIVED FUNCTIONS. 

41. Limiting Values. — The rules for differentiation 
have been deduced, in Chapter II, in accordance with the 
method of infinitesimals explained in Chapter I. We shall 
now deduce these rules by the method of limits. 

The limit, or limiting value of a function, is that value 
toward which the function continually approaches, till it 
differs from it by less than any assignable quantity, while 
the independent variable approaches some assigned value. 
If the assigned value of the independent variable be zero, 
the limit is called the inferior limit ; and if the value be 
infinity, it is called the superior limit. 

42. Algebraic Illustration. — Take the example, 

■ v = IT* 

and consider the series of values which y assumes when x 
has assigned to it different positive values. When x = 0, 
y = 1, and when x has any positive value, y is a positive 
proper fraction ; as x increases, y decreases, and can be made 
smaller than any assignable fraction, however small, by giv- 
ing to x a value sufficiently great. Thus, if we wish y to be 
less than 100 ^ 000 , we make x = 1000000, and get 

1 

y ~ iT"Toooooo 

which is less than 100 ^ 000 . If we wish y to be less than 
one-trillionth, we make x = 1000000000000, and the re- 
quired result is obtained. We see that, however great x 
may be taken, y can never become zero, though it may be 



(0 TRIGONOMETRIC ILLUSTRATION. 

made to differ from it by as small a quantity as we please. 

Hence, the limit of the function is zero when x is 

infinite. 

We are accustomed to speak of such expressions thus : 
" When x is infinite, y equals zero." But both parts of this 
sentence are abbreviations: "When x is infinite" means, 
" When x is continually increased indefinitely" and not, 
"When x is absolute infinity ;" &nd "y equals zero" means 
strictly, "y can be made to differ from zero by as small a 
quantity as we please." Under these circumstances, we say, 
" the limit of y, when x increases indefinitely, is zero." 

43. Trigonometric Illustration. — An excellent exam- 
ple of a limit is found in Trigonometry. To find the values 

of - — - and — ~ , when 6 diminishes indefinitely. Here 
tanfl • J 

we have 

- = cos 6 ; and when 6 = 0, cos = 1. 

tan0 

Hence, if 6 be diminished indefinitely, the fraction - — - 

will approach as near as we please to unity. In other 

words, the limit of z — B , as 6 continually diminishes, is 
tan C7 

unity. We usually express this by saying, " The limit of 

sin . „ "sin 6 „ 

- — - , when = 0, is unity ;" or, z — B = 1, when 6 = 0;" 
tan Q J tan 

that is, we use the words " when 6 = 0" as an abbreviation 

for " when 6 is continually diminished toioard zero." 

Since z — z = 1> wnen d = 9 

VdliO 

we have also -. — s == 1, when 6 = 0, 
sin 6 



TRIGONOMETRIC ILLUSTRATION. 61 

It is evident, from geometric considerations, that if 6 be 
the circular measure of an angle, we have 

tan d > 6 > sin 6 ; 

tan d ^ ■ . 

7JEi > ST9 >15 

but in the limit, t. &, when = 0, we have 

tan_0 _ 
sin (9 ~~ ' 

and therefore we have, at the same time, 

6 , sin © 

= 1, and .-. —J- = 1, 



sin 

which shows that, in a circle, the limit of the ratio of an arc 
to its chord is unity. 

"sin 6 
In the expression, — — = 1, when 6 = 0," it is evident 

that — — is never equal to 1 so long as 6 has a value dif- 
ferent from zero ; and if we actually make = 0, we render 

A1 . sin 6 ■ i * mu 4. « i-i sin 

the expression — — meaningless.* That is, while —^— 

approaches as nearly as we please to the limit unity, it never 
acti'cdly attains that limit. 

If a variable quantity be supposed to diminish gradually, till it be 
less than anything- finite which can be assigned, it is said in that state 
to be indefinitely small, or an infinitesimal ; the cipher is often used 
as an abbreviation to denote such a quantity, and does not mean abso- 
lute zero ; neither does go express absolute infinity. 

Rem. — The student may here read Art. 12, which is applicable to 
this method as well as to that of infinitesimals, which it is not neces- 
sary for us to insert again. 

* See Todhunter's Dif. Cal., p. 6. 



02 DERIVATIVES. 

44. Derivatives. — The ratio of the increment of u to 

A i)t 

that of x, when the increments are finite, is denoted by — ; 

the ratio of the increment of u to that of x in the fo'mtf, 

i e. 9 when both are infinitely small, is denoted by -=- , and 
is called the derivative * of u with respect to x. 

Thus, let u =f(x) ; and let x take the increment h 
(= Ax), becoming x -f- 7i, while u takes the corresponding 
increment Aw ; then we have, 

u -f- Au = f(x + h) ; 

therefore, by subtraction, we have 

Au==f(x + h)-f(x); 

and dividing by h (= Ate), we get 

Au _f(x + h)—f(x) 



Ax h 



a) 



It may seem superfluous to use both h and Ax to denote the same 
thing, but in finding the limit of the second member, it will sometimes 
be necessary to perform several transformations, and therefore a sin- 
gle letter is more convenient. In the first member, we use Ax on 
account of symmetry. 

The limiting value of the expression in (1), when tl 
is infinitely small, is called the derivative of %i or 
f(x) with respect to x, and is denoted by f (x). 

Therefore, passing to the limit, by making h diminish 
indefinitely, the second member of (1) becomes /' {x), and 

the first member becomes, at the same time, -y- ; hence we 
have 

* Called also the derived function and the differential coefficient. 



DIFFERENTIAL COEFFICIENT. 63 

45. Differential and Differential Coefficient. 

Let u = f(x) ; then, as we have (Art. 44), 

£=/'(*>> ' 

we have du = clf(x) = f (x) dx, 

where dx and du are regarded as being infinitely small, and 
are called respectively (x\rt. 12) the differential of x and the 
corresponding differential of u. 

f (x), which represents the ratio of the differential of the 
function to that of the variable, and called the derivative of 
f(x) (Art. 44), is also called the differential 'coefficient of 
f(x), because it is the coefficient of dx in the differential 
of/(*). 

Some writers * consider the symbol — only as a whole, and do not 

dx 

assign a separate meaning to du and dx ; others,! w ho also consider 

du 
the symbol -- only as a whole, regard it simply as a convenient nota- 

(XX 

tion to represent ^ , and claim that du and dx are each absolutely zero. 

46. Differentiation of the Algebraic Sum of a 
Number of Functions. 

Let y = au + bv -f- civ + z -f etc., 

in which y, u, v, w, and z are functions of x. Suppose that 
when x takes the increment h (= Ax), y, u, v, w, and z 
take the increments Ay, Au, Av, Aio, Az. Then we have, 

y + Ay = a (u + Au) -\-b(v-\- Av) + c (w -f Aiv) + (z -f A2) + etc. 
/. Ay = a Au -\- b Av + c Aw + Az -\- etc. 
Dividing by h or Ax, we have 

* See Todhunter's Dif. Cal., p. 17 ; also De Morgan's Calculus, p. 14, etc. 
t See Young's Dif. Cal., p. 4. 



64 DIFFERENTIATION OF A PRODUCT. 

Ay Au , , Av ' Aw , Az , 

-^ = a h # h c h • h etc., 

A£ A# A# Ax Ax 

which becomes in the limit, when h is infinitely small 
(Art. 12), 

dy du t 7 dv , dw clz , . . , H . . 

^ = a ^ + *5S + C S" + S + etc * (see Art 14) ' 

47. Differentiation of the Product of two Func- 
tions. 

Let y = ?«v, where w and v are both functions of x, and 
suppose Ay, Au, Av to be the increments of y, u. v corre- 
sponding to the increment Ax in x. Then we have 

. y -f Ay == (u + Au) (v + Av) 

= uv 4- « Av -f- v Aw + Au Av. 

.*. Ay = u Av + v A?£ + Aw Av ; 

Aw Av Au Au 
or, -£ = u - \- v \- -- Av. 

Ax Ax Ax Ax 

Now suppose Ax to be infinitely small, and 
Ay Av Au 

Ax' Ax' ~AZ 9 
become in the limit, 

dy dv du 

dx ' dx' dx 

Also, since Av vanishes at the same time, the limit of the 
last term is zero, and hence in the limit we have 

dy dv du /Q A . n _ . 

dx = u Tx + v Tx ( SeeArt - 16 -) 

lfr«l/ (.£•*< (J/Jy 

It can easily be seen that, although the last term vanishes, the 
remaining terms may have any finite value whatever, since they con- 
tain only the ratios of vanishing quantities (see Art. 9). For example, 

— = '- when x = ; but by canceling x we get — = a. But the 
x x 

expression — x x, which equals - x when x = 0, becomes - x = 

when x = 0. 



DIFFERENTIATION OF A PRODUCT. ()5 

Otherwise thus: 

Let f(x) (x) denote the two functions of x, and let 

u =f(x) 0(a). 

Change x into x -f h, and let u + &u denote the new 
product; then 

u -j- Au = f(x -f /*) (x + 7i) 

A*£ _ /( a + /Q ft (^ + /Q — /(a:) (x) _ 
A# " A 

Subtract and add f(x) (x + //), which will not change 
the value, and we have 

^ = fix + ft) -m {x + h) +f{x) tfe+iiriW. 

l\X Iv II 

Now in the limit, when h is diminished indefinitely, 

» (8 + ^-^ > = »'(») (Art 44); 

and ${x + h) = (.t) ; 

therefore, g = /' (a;) (a) + / (a?) 0' (a), 

which agrees with the preceding result. 

48. Differentiation of the Product of any Number 
of Functions. 

Let y = uvw, 

u, v, w being all functions of x. 

Assume z = viv ; 

then y = uz, 



66 DIFFERENTIATION OF A FRACTION. 

and by Art. 47 we have 

dy _ udz zdu 
dx dx dx 

Also, by the same Article, 

dz vdiv ivdv 

dx"" dx dx ' 

hence, by substitution, we have 

dy diu , dv du /0 A , „„, x 

-^ — w# - — \- uw -y + w-y. (See Art. 17.) 
«.t dx dx dx v . ' 

The same process can be extended to any number of 
functions. 

49. Differentiation of a Fraction. 

T w 

Let y = ■-• 

Then we shall have 

w + Aw 
y.+.Ay = 

Ay = 



v + Av ' 




?£ + Au 


u 


v + (/v 


V 


v Au — u 


A?; 









v 2 + v Ay 


Dividing by 


Ax and 


passing 


; to the limit, 






dy _ 


du 
dx 


dv 
dx 



dx v 2 

(since vA# vanishes). (See Art. 18.) 

Cor. — If u is a constant, we have 

udv 
dy _ dx 

dx ~ v 2 



DIFFERENTIATION OF ANY POWER. 67 

50. Differentiation of any Power of a Single Va- 
riable. 

1st. When n is a positive integer. 

Let y = x n ; 

then we have y + by = (x + h) n ; 

therefore, by — nx n ~ l h H K —- — ; x n ~ 2 h 2 + etc. . . , h\ 

1 • /£ 

Dividing by h or bx, we get 

by , n (n — 1) „ , 7 , 

Arc 2 

Passing to the limit, we have 

^ = nx n ~\ (See Art. 19, 1st.) (1) 

2d. When n is a positive fraction. 

m 

Let y = u\ 

where u is a function of x ; then 





if = u 1 *, 




d(yY = d(u m ); 


s by (i), 


„ , dij m , *?w 

n yn-l^L _ mu m-\ 

v ax ax 




dy m u m ~ l du 
dx ~ n y n ~ l dx 



3d. When ri is a negative exponent, integral or 
fractional. 

Let y = ur* ; 

then y = — , 



68 DIFFERENTIATION OF LOGARITHMS, 

and by Art. 49, Cor., we have 

rhi, 

(3) 



dy 

dx 


nu n ~ l du „ , du , A , - ft t v 

= — — = ww-"- 1 -y- (Art. 19, 3d). ( 

u ln dx dx v ? / v 


51. Differentiation of log x. 


Let 


# = log x\ 


therefore, 


y + Ay = log (x + ^), 


and 


A?/ = log (x -f h) — - log x 




^m —('+j| 




- ? f* 7 ^ 2 *' "i • 

\# # 2 ^ 3 ' / ' 


therefore, 


Aw /l 7* \ 

-f- = m{ + etc. 1 ; 

Ax \x x? 1 


therefore, 


passing to the limit, we get 




dy m 1 

-f = — or - 

CIX X X 



(according as the logarithms are not or are taken in the 
Naperian system. See Art. 20). 

52. Differentiation of a x . 

Let y == a x . 

Proceeding exactly as in Art. 21, we get 

■Q = — log a or a x log a (Art. 21). 

CtX tit 

53. Differentiation of sin ac. 

Let y = sin ^ ; 

therefore y + Ay — sin (x + ^) ; 

hence, Ay = sin (# -f- h) — sin a?. 



DIFFERENTIATION OF A COSINE. 69 

But from Trigonometry, 

. . . -. . A + B . A-B 

sm A — sm B = 2 cos 









uo 2 


"" 2 


^ 


= 


sin 


(x + h) 


— sin a; 




z= 


2 cos ix + 


/A , A 
2J Sm 2' 




— 


cos 


(*+'*) 


sm- 
h 



hence, 

2 

By Art. 43, when h is diminished indefinitely, the limit of 

. h 
sm- 



1 ; also, the limit of cos ix + ^ ) = cos x. 



2 



Therefore, -v- = cos x. (See Art. 24.) 

ax v 

54. Differentiation of cos x. 

Let y =. cos $c ; 

therefore, y -\- Ay = cos (a; -{- h) ; 

hence, A?/ = cos (x -{- h) — cos x 

7i\ . li 



==-««*(*+.$ ring, 



. /A + B\ A-B 



because cos A — cos B = — 2 sin ( — - — ) sin 

sin 



Therefore, ~ = — sin Jo; + ^1 



* 

* 

2 



70 COMPARISON OF THE TWO METHODS. 

Hence, in the limit, 

-^ = — sin x. (See Art. 25.) 
dx 

Of course this differentiation may be obtained directly 
from Art. 52, in the same manner as was done in the 2d 
method of Art. 25. 

Since tan x, cot x, sec x, and cosec x are all fractional 
forms, we may find the derivative of each of these functions 
by Arts. 18 or 49, from those of sin x and cosic, as was done 
in Arts. 26, 27, 28, and 29; also, the derivatives of vers x 
and covers x, as well as those of the circular functions, may 
be found as in Arts. 30, 31, 33 to 40. 

Prom the brief discussion that we have given, the student 
will be able to compare the method of limits with the method 
of infinitesimals ; he will see that the results obtained by 
the two methods are identically the same. In discussing by 
the former method, we restricted ourselves to the use of 
limiting ratios, which are the proper auxiliaries in this 
method. It will be observed that, in the former method, 
very small quantities of higher orders are retained till the 
end of the calculation, and then neglected in passing to the 
limit; while in the infinitesimal method such quantities 
are neglected from the start, from the knowledge that they 
necessarily disappear in the limit, and therefore cannot 
affect the final result. As a logical basis of the Calculus, 
the method of limits may have some advantages. In other 
respects, the superiority is immeasurably on the side of the 
method of infinitesimals. 



CHAPTER IV. 



SUCCESSIVE DIFFERENTIALS AND DERIVATIVES. 



55. Successive Differentials. — The differential ob- 
tained immediately from the function is the first differential. 
The differential of the first differential is the second differ- 
ential, represented by d 2 y, d 2 u, etc., and read, "second 
differential of y 9 " etc. The differential of the second dif- 
ferential is the third differential, represented by d B y, d 3 u, 
etc., and read, " third differential of y," etc. In like man- 
ner, we have the fourth, fifth, etc., differentials. Differen- 
tials thus obtained are called successive differentials. 

Thus, let AB be a right line 
whose equation is y = ax -f- b ; 
.-. dy = adx. Now regard dx as 
constant, i. e., let x be equicres- 
cent;* and let MM', M'M", and 
M"M"' represent the successive 
equal increments of x, or the dx's, 
and R'P', K"P", R'"P" the corre- 
sponding increments of y, or the 

fly's. We see from the figure that E'P' = R"P" = R'"F" ; 
therefore the dy's are all equal, and hence the difference 
between any two consecutive dy's being 0, the differential 
of dy, i.e., dry — 0. Also, from the equation dy = adx we 
have d 2 y = 0, since a and dx are both constants. 

Take' the case of the parabola if — 2px (Fig. 7), from 
pdx 

y 




Fig. 6 



which we get dy 



Regarding dx as a constant, we 



* When the variable increases by equal increments, i. e., when the differential i& 
constant, the variable is called an eqidcrescent variable. 



72 



EXAMPLES. 



have MM', M'M", M"M"' as the successive equal increments 
of x, or the dx's ; while we see from 
Fig. 7 that RT', R'T", R'"F", or 
the dy% are no longer equal, but 
diminish as we move towards the 
right, and hence the difference be- 
tween any two consecutive dy's is a 
negative quantity (remembering that 
the difference is always found by 
taking the first value from the second. 
See Art. 12). Also, from the equa- 

7) 

tion dy = - dx we see that dy varies inversely as y. 

The student must be careful not to confound d 2 y with 
dy 2 or d(y 2 ): the firsij is "second differential of y.\ n the 
second is "the square of dy;" the third is the differential 
of y 2 , which equals 2ydy. 




EXAMPLES. 

1. Find the successive differentials' ofy = x 5 . 
Differentiating, we have dy = 5a; 4 dx. Differentiating 

this, remembering that d of dy is d 2 y and that dx is con- 
stant, we have d 2 y = 20a; 3 dx 2 . In the same way, differen- 
tiating again, we have d 2 y = 60x 2 dx 3 . Again, d*y = 120:t dx*. 
Once more, d s y = 120dx 5 . If we differentiate again, we 
have d 6 y = 0, since dx is constant. 

2. Find the successive differentials of y = 4:X 3 —3x 2 + 2x. 

I dy = {l%x 2 — 6x + 2) dx ; 
Ans. \ d 2 y = (Ux — 6) dx 2 ; 
( dh) = 24^ 3 . 

3. Find the first six successive differentials ofy = sin #. 

f dy = cos £ afo: ; d 2 y = — sin x dx 2 ; 

Ans. < d B y = — cos x dx 3 ; d*y = sin x dx* ; 

( d 5 y = cos x dx 5 ; d 6 y = — sin a; dx 6 . 



EXAMPLES. 



73 



4. Find the first six successive differentials of y = cos x. 

(dy = — sin x dx ; d 2 y = — cos x dx 2 ; 

Ans. i d 3 y = sinoj dx 3 ; d*y — cos a; ofa?*; 

( #?/ = — sin a; dx 5 ; d 6 y = — cos x dx\ 

5. Find the fourth differential of y = x". 

Ans. dhj — n (n — I) (n — 2) (n — 3) x^dx 4 . 

6. Find the first three successive differentials of y = a x . 

i dy = a x log a dx ; 

AmA dhj = a x fog 2 adx 2 ; 

( d 3 y = a x \o^a dx 2 . 

7. Find the first four successive differentials of y = logic. 



Ans. ' 



dy = 



dx 



d 2 y 



dx 2 



dhj = - 



Mx 



,3 > 



7 , 6dx* 

d *y= -—*-' 



8. Find the first four successive differentials of y = 2a\fx, 

adx 



Ans. 






70 adx 2 

&y = - —j ; 

yx 2x* 



3adx? 
~±x% 



dhj = - 



15 adx 4 
8z* 



9. Find the first four successive differentials of 
y = log (1 + x) in the common system. 



Ans. - 



, mdx 

dii = - 

J 1 + x 



rPy = 



2 mdx' 3 



d 2 y = - 
dhj = - 



mdx 2 

(1 + s)* 5 

6m dx 4 



10. Find the fourth differential of y = e x . 

Ans. d 4 y = e x dx*. 
4 



74 SUCCESSIVE DERIVATIVES. 

.. 56. Successive Derivatives.— A first derivative* is 
the ratio of the differential of a function to the differential 
of its variable. Tor example, let 

y — ^ 

represent a function of x. Differentiating and dividing by 
dx, we get 

d 4 = Gafi. (1) 

dx v J 

The fraction -j- is called the first derivative of y with 

respect to x, and represents the ratio of the differential of 
the function to the differential of the variable, the value of 
which is represented by the second member of the equation. 

Clearing (1) of fractions, we have 

dy = 6x 5 dx ; 

dv 
hence, —- or 6x 5 is also called the first differential coefficient 

of y with respect to x 9 because it is the coefficient of dx. 

A second derivative is the ratio of the second differential 
of a function to the square of the differential of the variable. 
Thus, differentiating (1) and dividing by dx, we get (since 
dx is constant, Art. 55), 

8 = 3o * 4 ' <*> 

either member of which is called the second derivative of y 
with respect to x. 

A third derivative is the ratio of the third differential of 
a function to the cube of the differential of the variable. 
Thus, differentiating (2) and dividing by dx, we get 



* See Arts. 44 and 45. 



DIFFERENTIAL COEFFICIENTS. 75 

either member of which is called the third derivative of y 
with respect to x. 

In the same way, either member of 

g = 360^ (4) 

is called the fourth derivative of y with respect to x, and 
so on. 

Also, -~, -=■¥. J- 9 , J, etc., are called respectively the 
dx dx?- dx s dx* r J 

first, second, third, fourth, etc., differential coefficients of 

y with respect to x, because they are the coefficients of dx, 

dx 2 , dx 3 , dx*. etc., if (1), (2), (3), (4), and so on, be cleared 

of fractions. 

In general, if y =f(x), we have 

i =^ l) = f ' {x) ( Art - 45 ); ••• d v = /'(*)<**■ 

etc. = etc. = etc. .*. etc. = etc. 

fry = rf/^>(^ 



/ (M) (a;) ; .\ d n y = /<»> («) dx\ 



That is, the first, second, third, fourth, etc., derivatives 
are also represented by /' (x), f (x), f" (x), f iV (x), etc. 



76 GEOMETRIC REPRESENTATION. 

dv 
Strictly speaking, -^ or/' (x) are symbols representing 

the ratio of an infinitesimal increment of the function to the 
corresponding infinitesimal increment of the variable, while 
the second member expresses its value. For example, in 
the equation y = ax*, we obtain 

-j- or /' (x) is an arbitrary symbol, representing the value 

cix 

of the ratio of the infinitesimal increment of the function 
(ax 4 ) to the corresponding infinitesimal increment of the 
variable (x), while ±ax» is the value itself. It is usual, 
however, to call either the derivative. 

56^. Geometric Representation of the First De- 
rivative. — Let AB be any plane curve 
whose equation is y-=f(x). Let P 
and P' be consecntive points, and PM 
and P'M' consecutive ordinates. The 
part of the curve PP', called an ele- 
ment* of the curve, does not differ , / 
from a right line. The line PP' pro- T 

longed is tangent to the curve at the 




Fig. 8. 



point P (Anal. Geom., Art. 42). Draw PR parallel to XX', 
and we have 

MM' = PR = dx, and RP' = dy. 

Denote the angle CTX by a, and since CTX = P'PR, we 
have 

dy 

tan a = — c -- 
dx 

And since the tangent has the same direction as the curve 

* In this work, the word " element' 1 will be used for brevity to denote an "in- 
finitesimal element." 



EXAMPLES. 77 

at the tangent point P, « will also denote the inclination of 
the curve to the axis of x. 

Hence, the first derivative of the ordinate of a curve, 
at any -point, is represented by the trigonometric tan- 
gent of the angle which the curve at that point, or its 
tangent, makes with the axis of x* 

In expressing the above differentials and 'derivatives, we 
have assumed the independent variable x to be equicrescent 
(Art. 55), which we are always at liberty to do. This 
hypothesis greatly simplifies the expressions for the second 
and higher derivatives and differentials of functions of x, 
inasmuch as it is equivalent to making all differentials of x 
above the first vanish. Were we to find the second deriva- 
tive of y with respect to x, regarding dx as variable, we 
would have 

&y _ d l^y\ __ dxd 2 y — dy cftx 
~M ~ dx \dx) ~~ -daT ' 

cPu 
which is much less simple than the expression -~, obtained 

by supposing dx to be constant. 



EXAMPLES. 

1. Given y = x n , to find the first four successive deriv- 
atives. 

dx = nX ' 

g = »(n-l)(n- 2)*-»; 

^ = n(n - I) (n - 2) (n - S)**-*. 



78 EXAMPLES. 

If n be a positive integer, we have 

g = »(»-l)(«-2)....3.2.1. 

and all the higher derivatives vanish. 

If n be a negative integer or a fraction, none of the suc- 
cessive derivatives can vanish. 

d 4 u 
2. Given y = a? log x ; find y|- 

~ = 3a: 2 log x + x % ; 

-^| = 6x log a; + Sx + 2rr = 6a: log a; + 5a; ; 

? s = 61ogx + 6 + 5. ** = ! 

da 3 & da 4 x 

It can be easily seen that in this case all the terms in the 
successive derivatives which do not contain log x will dis- 
appear in the final result; thus, the third derivative of a 2 is 
zero, and therefore that term might have been neglected ; 
and the same is true of 5x, its second derivative being zero. 

■ 1 + x ,, , d 5 y 240 

d^v 

4. y = e ax ; prove that j~ = ah ax . 

5. y = tan x ; find the first four successive derivatives. 

dy 9 

-£- = sec 2 x ; 
dx 

—4 = 2 sec 2 x tan # ; 
da^ 

-=-| = 6 sec 4 a; — 4 sec 2 a; ; 

-y^ = 8 tan a; sec 2 a; (3 sec 2 a; — 1). 
dx* v ' 



EXAMPLES. 79 

6. y =z leg sin x; prove that —^ = 2 cot a; cosec 2 #. 

7. # 2 = ^ ; find ||- 







ty _P. 
dx y' 




= - £) = ■ 


dy p 

13 dx Py (. c 

Y- = f ( since - 

y 2 y % V c 


d s y 

dx^ ~ 


" dx \ yV 


3^2^ 3/^ Q c 
1 y dx _ r y ?>p* 

y 6 y* ' y 5 


8. y: 


= of ; prove 


that 



C —^ — x x (1 + log xf -f x*- 1 . 



9. a 2 ?/ 2 + # 2 # 2 == a 2 b ; prove that ~ = z—« 

^ 3 4.1. ^ 4 # 24 

10. ?/ = ; prove that -=^ = 



11. f- = sec 2x ; prove that 3/ -f y| = 3?/ 5 . 

12. ?/ = e - * cos x ; prove that % -f- v| = 0. 

13. y = a? log (a; 8 ) ; prove that v| = — 

ax x 



80 EXAMPLES. 

14. y = a? ; prove that 

clhj = 6 {dxf + 18zdzd 2 z -J- 3aftZ 3 #, 
when a; is not equicrescent. 

15. y = f(z); prove that 

flfy = /'" (a;) (dx)* + 3/" (a;) dx&x +/' (a;) fc, 
when x is not equicrescent. 

16. y = e x ; prove that 

#2/ = e* (da?) 3 + Se x dz d 2 z + e x tf s z. 



CHAPTER V. 

DEVELOPMENT OF FUNCTIONS. 

57. A function is said to be developed, when it is 
transformed into an equivalent series of terms following 
some general law. 

For example, 

y = (a + xf, 

when developed by the binomial theorem, becomes 

y = a 4 -f ±a 3 x -f 6a?x 2 + ±ax z -f x*, 
which is & finite series. Also, 

1 + x 

may be developed by division into the infinite series, 

y = 1 -f 2x + 2x* + 2r 3 -f etc., 

in which the terms are arranged according to the ascending 
powers of x, each coefficient after the first term being 2. 

One of the most useful applications of the theory of suc- 
cessive derivatives is the means it gives us of developing 
functions into series by methods which we now proceed to 
explain. 

MACLAURIN'S THEOREM. 

58. Maclaurin's Theorem, is a theorem for developing 
a function of a single variable into a series arranged 
according to the ascending powers of that variable, with 
constant coefficients. 



82 MACLAURIN'S THEOREM. 

Let y=/» 

be the function to be developed ; and assume the develop- 
ment of the form 

y = f(x) = A + Bx + Cx 2 + Dx 5 + Ex* + etc., (1) 

in which A, B, C, D, E, etc., are independent of x, and 
depend upon the constants which enter into the given func- 
tion, and upon the form of the function. It is now re- 
quired to find such values for the constants A, B, C, etc., 
as will cause the assumed development to be true for all 
values of x. 

Differentiating (1) and finding the successive derivatives, 
we have, 

g| = B + 20x + SDx* + ±Ex* + etc., (2) 

^| = 2C + 2 • ZDx + 3 . ±Ex* + etc., , (3) 

CIX 

^ = 2 • ZD + a • 3 • 4MX + etc., (4) 

g = 2.3.4i7+ete., (5) 

Now, as ^4, 5, (7, etc., are independent of x, if we can 
find what they should be for any one value of x, we shall 
have their values for all values of x. Hence, making x = 
in (1), (2), (3), etc., and representing what y becomes on 

this hypothesis by (y) ; what ~- becomes by \-~\ ; what 
Y~ becomes by ( -- "A ; and so on ; we have, 
(y)=A; /. A = (y). 



MACLAURIN'S THEOREM. 83 

Substituting these values in (1), we have, 

»-/(-) = w + ffli + fflA+ffllnfi 

which is the theorem required. 

Hence, by Maclauriirs Theorem, we may develop a func- 
tion of a single variable, as y = f (x), into a series of terms, 
the first of which is the value of the function when x = 0; 
the second is the value of the first derivative of the function 
when x = into x ; the third is the value of the second 

x 2 
derivative when x = into -^- , etc. ; the nth. term is 



x n 



. . n \dx a f 



1.2-3 

We may also use the following notation for the function 
and its successive derivatives : fix), f (x), f" (x), f" (x), 
f**(x), etc., as given in Art. 56, and write the above 
theorem, 

».= /(*) r /(o) + f'M \ + /" (0) jf 2 + /'" (0) r ^ 

+ /"(0) r7 ^— 4 +etc, (7) 

in which /(0), /' (0), /" (0), /'" (0), etc., represent the 
values which f(x) and its successive derivatives assume 



MA CLA URIN'S THEOREM. 



when x — 0. We shall use this notation instead of 

i . > 1% > e ^ c -> ^ or ^ e sa ^- e °^ DreYr ty- 

This theorem, which is usually called Maclaurin's Theorem, was 
previously given by Stirling in 1717 ; but appearing first in a work on 
Fluxions by Maclanrin in 1742, it has usually been attributed to him, 
and has gone by his name. Maclaurin, however, laid no claim to it, 
for after proving it in his book, he adds, " this theorem was given by 
Dr. Taylor." See Maclaurin's Fluxions, Vol. 2, Art. 751. 

To Develop y = (a + x) 6 



Here 
hence, 



f(x) = (« + *)•; 

/(0) = a\ 
f(x) = 6(a + xf; 
f(0) = 6a\ 
f"{x) = 5.6(a + a;)«; 
/"(0) =5-6^. 



f"(x) =4-5 
/"'(0)=4-5 

/' lv W = 3.4 
f (0) = 3 • 4 

/"(*) = a- 3 

/ v (0) = 2.3 
/-■(«) =1.8 
f(0) =l.fl 

Substituting in (7), we have, 



6(a.+ xf; 



6{a + *) 8 s 

6a 2 . 

5 • 6 (a -f- #) ; 

6a; 

5- 6; 

5-6; 



y=(a + xf 
+ 3.4.5.6 



a 6 4- 6a 5 # -h 5 
ffl¥ 2-3. 4.5 



6^^ + 4.5. 



+ 



+ 



1-2- 3 
1-2-3.4. 5- 6a 8 



1.2.3.4 ' 1.2.3.4.5.6 ' 1.2.3.4.5.6 
= a 6 + 6a 5 x + loaW + 20« 3 £ 3 + 15aV + 6aa; 5 + .? 6 , 



THE BINOMIAL THEOREM. 85 

which is the same result we would obtain by the binomial 
theorem. 

THE BINOMIAL THEOREM. 

59. To Develop y — {a + x) n . 

Here f(x) = (a + x) n \ 
hence, /(0) == a\ 

f (x) = n (a + a)"- 1 ; 

/' (0) = na n ~\ 

f" (x) = n (n — 1) (a + #)*" 2 . 

/" (0) = n (n - 1) a"" 2 . 

/'" (a) =w(ffl~l)(»-2)(a + z)"- 3 ; 

/'" (0) = n (n — \)(n — 2) a n ~\ 

f" (x) = n(n — 1) (n — 2) (?* — 3) (a + ^) n_4 ; 

f*(0) = n (n — 1) (h — 2) (» — 3) a"S etc. 

Substituting in (7), Art. 58, we have, 

n (n — 1) a n - 2 x* 
y = (a + a)" = a B + ^ _1 £ H — -, — k 

n (n — 1) (n — 2) a n ~ z x z 
+ l"T~3 

w in — 1) (w — 2) (n — 3) «"" 4 ^ 4 
+ ^ 8.84 + etC " 

Thus the truth of the binomial theorem is established, 
applicable to all values of the exponent, whether positive 
or negative, integral or fractional, real or imaginary. 

60. 1. To Develop y — sin x. 

Here f(x) = sin x ; hence, /(0) = 0. 

" f'-(x) = cosx; " /'■(()) = 1. 

" /" (x) = — sin x ; " /" (0) = 0. 



8G THE LOGARITHMIC SERIES. 

Here /'" (x) = — cos x; hence,/'" (0) = — 1. 
" f (x) = sin x ; " f r (0) = 0. 

f (x) = cos x; " f (0) = 1. 

Etc., etc. Etc., etc. 

x^ x^ 

Hence, y = sin x = x - j-^ + 1 .^. z ^. 5 

+ etc. 



1.2.3.4.5.6-7 



2. To Develop y = cos x. 

Ans. y = cos x = 1 — - — - 4- 



1-2 ' 1.2.3.4 



1.2.3.4.5.6 ' 1.2.3.4.5.6.7.8 



— etc. 



The student will observe that by taking the first derivative of the 
series in (1), we obtain the series in (2), which is clearly as it should 
be, since the first derivative of sin x is equal to cos x. 

Since sin (— x) = — sin x, from Trigonometry we might have 
inferred at once that the development of sin x in terms of x could con- 
tain only odd powers of x. Similarly, as cos (— x) = — cos x, the 
development of cos x can contain only even powers. 

By means of the two formulas in this Article we may 
compute the natural sine and cosine of any arc. For exam- 
ple, to compute the natural sine of 20°, we have x = arc of 

20° = ^ = .3490652, which substituted in the formulas, 

gives sin 20° = .342020 and cos 20° = .939693. 

THE LOGARITHMIC SERIES. 

61. To Develop y — log (1 + x) in the system in 
which the modulus is m. 

Here f{x) = log (1 + x); 'hence, /(0) = 0. 



THE LOGARITHMIC SERIES. 87 



m 

1 +~x)< 



Here/" (x) = — n , , 2 ; hence, /" (0) = — m. 



" f (*) = - ^nflf 5 " /- (0) = - . • 2 . 8m. 
Etc. Etc. 

Substituting in (7), Art. 58, we have, 
y = log {1+x) = m (x— ^ + i« 3 — i^ 4 + i^ 5 — etc.), (1) 
which is called the logarithmic series. 

Since in the Naperian system m = 1 (see Art. 20, Cor.), 
we have, 

ovj /y3 /v4 /y>5 

y = log (1 + x) = x - - + g- - j + g- — etc. (2) 

which is called the Naperian logarithmic Series. 

This formula might be used to compute Naperian loga- 
rithms, of very small fractions j but in other cases it is 
useless, as the series in the second number is divergent for 
values of x > 1. We therefore proceed to find a formula 
in which the series is convergent for all values of x ; i. e., in 
which the terms will grow smaller as we extend the series. 

Substituting — x f or x in (2), we have, 

/y2 W& n/A eY*5 

log (1-*)=-^-?- ?-£-?- etc. (3) 

Subtracting (3) from (2), we have, 

i ,, i /. n 2£ 3 2x 5 2x* 

log (1 + x) - log (1 - x) = %x + -y + -y + -y + etc. 



or 



T Z 1 + ®\ ., / ^ ^ ^ \ /„V 

Mf^) = 5 T + 3 +5 +7 +ete l (4) 



88 CALCULATION OF LOGARITHMS. 

T , 1 1 H- x % -+ 1 

Let z = ^ — — r ; .*. -— — = — 

2z + 1 1 — a ^ 

Substituting in (4), we have, 

2 + l_rl 1 : 1 

g "7~ ~ L2z -f 1 + 3 (2s + l) 3 + 5 (2s + l) 5 

+ 7p7+iy* + etc *] 5 
or log(, + l) = log, + 2[^ I+ 3 W i- 1? 

^ s"(^ + iy + y^iTiy + ^' (5) 

This series converges for all positive values of z, and more 
rapidly as z increases. By means of it the Naperian loga- 
rithm of any number may be computed when the logarithm 
of the precedi?ig number is known. It is only necessary to 
compute the logarithms of prime numbers from the series, 
since the logarithm of any other number may be obtained 
by adding the logarithms of its factors. The logarithm of 1 
is 0. Making z = 1, 2, 4, 0, etc., successively in (5), we 
obtain the following 

Naperian or Hyperbolic Logarithms. 
lo g 2 = logl + 2(1 + 3^3-3 + 5^5 + ^ +9^9 + 



11. 3 11 



1 1 1 \ 

+ 13^3^3 .+ 15^5 + 1 7 . 317 + etc '/ > 



or, since log 1 = 0, 

,33333333 

,01234568 
.00082305 
log 2 = 2 ( .00006532 ) = 2 (0.34657359) == 0.69314718. 
.00000565 ' 
.00000051 
.00000005 



CALCULATION OF LOGARITHMS. 89 

log 3 = log 3 + sg + A, + ^ + A-, + ~ + etc.) 















= 1.09861228. 


log 


4 


= 2 log 2 








= 1.38629436. 


log 


5 


= log4 + 2g + - 


1 

;-9 3 + 


1 

5-9 5 


+ 1 


1 1 \ 

7.9 7 + 9.9 s + 7 

= 1.60943790. 


log 


6 


= log 3 -f- log 2 








= 1.79175946. 


log 


7 


= % 6 + 2 (— + 


l 
3-13 3 


+ 5 


1 
• 13 5 


+ ^3- 7 + etc.) 
= 1.94590996. 


log 


8 


= 3 log 2 








= 2.07944154. 


log 


9 


= 2 log 3 








. = 2.19722456. 


log 


10 


== log 5 -+- log 2 








= 2.30258509. 



In this manner, the Naperian logarithms of all numbers 
may be computed. Where the numbers are large, their 
logarithms are computed more easily than in the case of 
small numbers. Thus, in computing the logarithm of 101, 
the first term of the series gives the result true to seven 
places of decimals. 

Cor. 1. — From (1) we see that, the logarithms of the 
same number in different systems are to each other 
as the moduli of those systems; and also, that the 
logarithm of a number in any system is equal to the 
Naperian logarithm of the same number into the 
modulus of the given system. 

Cor. 2.— Dividing (1) by (2), we have 

Common log (1 + x) __ . * 

Naperian log (i + x) 

Hence, the modulus of the common system is equal 
to the common logarithm of any number divided by 
the JVaperian logarithm of the same number. 



90 EXPONENTIAL SERIES. 

Substituting in (6) the Naperian logarithm of 10 com- 
puted above, and the common logarithm of 10, which is 1. 
we have 

m = 2 30258509 = • 43429448190325182 ^ 6511289 • • • 

which is the modulus of the common system. (See Serret's 
Calcul Differentiel et Integral, p. 169.) 

Hence, the common logarithm of any number is 
equal to the JVaperian logarithm of the same number 
into the modulus of the common system, .43429448- 

Cor. 3. — Eepresenting the Naperian base by e (Art. 21, 
Cor. 2), we have, from Cor. 1 of the present Article, 

com. log e : Nap. log e (= 1) :: .43429448 : 1 ; 

therefore, com. log e = .43429448 ; 

and hence, from the table of common logarithms, we have 

e = 2.718281 + . 

EXPONENTIAL SERIES. 

62. To Develop y = a*. 

Here f(x) = a* ; hence, /(0) = 1. 

/' (x) = a x log a ; " f (0) = log a. 

« f" (x) = a* (log aY; « /" (0) = (log a)\ 

« /'" (*) = a* (log of ; « /"' (0) = (log a)K 
and the development is 

y - a* = 1 + log a j + log 2 a ~ + log 3 « j-^g 

+ log 4 a I -^-- + etc. (1) 



, EXPONENTIAL SERIES. 91 

Cor. — If a = e, the Naperian base, the development 
becomes 

x x 2 x s x* 



y = & = 1 + - + — ,+ - — + 



1 ' 1-3 ' 1-3-3 ' 1.2.3-4 

+ --irrb* + eto - (2) 

Putting x = 1, we obtain the following series, which en- 
ables us to compute the value of the quantity e to any 
required degree of accuracy : 

„ 111 1 



2 ' 2-3 ' 2.3.4 ' 2.3.4.5 

+ ---'2T i 7.T, + etc - 

= 2.718281828 + . 

63. To Develop y = tan -1 x. 

In the applications of Maclaurin's Theorem, the labor in 
finding the successive derivatives is often very great. This 
labor may sometimes be avoided by de\_ oping the first 
derivative by some of the algebraic processes, as follows : 

Here f(x) = tan -1 x; 
hence, /(0) = 0. 

= (by division) 1— x 2 -\- x i — x 6 -\- x 8 ; 
/'(0) = 1. 

f" (x) = — 2x + 4a: 3 — 6x 5 + 8a; 7 — 10a; 9 + etc. ; 
/"(0) = 0. 

f'"(x) = — 2 + 3- 4a; 3 — 5-6^ + 7-8a; 6 — etc.; 
f'"(0) = -2. 

f"(x) = 2.3-4o;-- 4.5.6a; 3 + etc.; 
/ iv (0) = 0. 



92 FAILURE OF MACLAURIN'S THEOREM^ 

f>(x) = 2-3.4 — 3.4.5.6^ + etc.; 
/ T (0) = 2-3.4. 

f l (x) = —2.3.4.5.6^ + etc.; 
/ vi (0) = 0. 
f™(x) = _2.3.4.5.6 -fete; 

/™(o) = —2.3.4.5.6. 

Substituting in (7) of Art. 58, we get 

/v3 /y5 /W 

«/ = tan -1 x = x — '~ -\- ~ + *=- + etc. 
o 7 

64. It sometimes happens in the application of Maclau- 
rin's Theorem that the function or some of its derivatives 
become infinite when x == 0. Such functions cannot be 
developed by Maclaurin's Theorem, since, in such cases, 
some of the terms of the series would be infinite, while the 
function itself would be finite. 

For example, take the function y = log x. Here we 
have 

f(x) = logrr; hence, /(0) = —00. 

/•■(*) = |; " /''(<>)"=<»• 

r(*) = -#i " /"(a) = -00. 

etc. . etc. 

Substituting in Maclaurin's Theorem, we have 

x x 

y = log x = — ao -\- co'- — co'— -f etc. 

JL 4/ 



Here we have the absurd result that log x = 00 for all 
values of x. Hence, y = log x cannot be developed by 
Maclaurin's Theorem. 

Similarly, y = cot x gives, when substituted in Maclau- 
rin's Theorem, 



TAYLOR'S THEOREM. 93 



X 

y = cot x = oo — oo - + etc. ; 

that is, cot x = oo for all values of x, which is an absurd 
result. Hence, cot x cannot be developed b}* Maclaurin's 
Theorem. 

Also, y — X* becomes, by Maclaurin's Theorem, 

y = x* = + oo x -f etc. ; 

that is, x* = oo for all values of x, which is an absurd 
result. 

Whether the failure of Maclaurin's Theorem to develop correctly is 
due to the fact that the particular function is incapable of any devel- 
opment, or whether it is simply because it will not develop in the 
particular form assumed in this formula, the limits of this book will 
not allow us to enquire. 



TAYLOR'S THEOREM. 

65. Taylor's Theorem is a theorem for developing a 
function of the sum of two variables into a series arranged 
according to the ascending powers of one of the variables, 
with coefficients that are functions of the other variable and 
of the constants. 

Lemma. — We have first to prove the following lemma: 
If we have a function of the sum of two variables x and y, 
the derivative will be the same, whether we suppose x to 
vary and y to remain constant, or y to vary and x to remain 
constant. For example, let 

u = (x + y)\ (1) 

Differentiating (1), supposing x to vary and y to remain 
constant, we have 



94 TAYLOR'S THEOREM. 

Differentiating (1), supposing y to vary and x to remain 
constant, we have 

$j=n(z + ,)**; (3) 

from which we see that the derivative is the same in both 
(2) and (3). 

In general, suppose we have any function of x + y, as 

u=f(x + y). (4) 

Let z = x + y ; (5) 

■•■ «=/(*). (6) 

Differentiating (5), supposing x variable and y constant, 
and also supposing y variable and x constant, we get 

dz . , dz 

T = l, and -7- = 1. 

dx dy 

Differentiating (6), we have 

du d f(z) ■. , x /a . , .„ . 

5 = "* =^<«>- ( SeeArt - 45 -) 



<??/ == /' (z) dz. 



du 
dx 



= ''<*)!=/' <*>( since J =4 



/'(*) (since | = l). 



And similarly, 




tfw _ f , , ,dz 
dy ~ 'J dy ' 


=/'( 


du 


du 


dx 


~ ty 



That is, the derivative of u with respect to x, y being 
constant, is equal to the derivative of u with respect 
to y, x being constant. 



METHOD OF TAYLOR'S THEOREM. ' 95 

66. To prove Taylor's Theorem. 

Let u' =f(x + y) be the function to be developed, and 
assume the development of the form 

v! = f(x + y) 

= A + By + Cif + Dif + Exf + etc., (1) 

in which A, B, C, etc., are independent of y, but are func- 
tions of x and of the constants. It is now required to find 
such values for A, B, C, etc., as will make the assumed 
development true for all values of x and y. 

Finding the derivative of u' 3 regarding x as constant and 
y variable, we have 

g-' = B + 2Cy + Wf + lEtf + etc. (2) 

Again, finding the derivative of u', regarding x as varia- 
ble and y constant, we have 

du' dA clB , dC 9 dD ' 

By Art. 65, we have -— = -=—: therefore, 
J dy dx ' 

B + 2Cy + Z Df + lEf+ete. = g + g f + g ^ + g ,3 

+ etc. (4) 

Since (1) is true for every value of y, it is true when 
y = 0. Making y = in (1), and representing what i*' 
becomes on this hypothesis by u, we have 

u=f(x) = A. (5) 

Since (4) is true for every value of y, it follows from the 
principle of indeterminate coefficients (Algebra) that the 
coefficients of the like powers of y in the two members 
must be equal. Therefore, 



96 TAYLOR'S THEOREM. 



n- dA 


• • 


B = %' imm ^> 


dx 


• 


p 1 d 2 u 


3D-^ 


•'• 


j^ 1 dht 


dx 


" 1-2-3 dx*' 


A IP dD 

* E =dx-> 


.*• 


F 1 dht 
"" 1-2.3-4' dx*' 



Substituting these values of A, B, C, D, etc., in (1), we 
have 

u * - f(x , y) _ u + d ±y s d2u y 2 , d%u y z 

dht «/ 4 

+ &i 1^374 + etc - < 6 > 

Or, using the other notation (Art. 56), we have 

+/ " (a;) r&"4 + etc :' (7) 

which is Taylor's Theorem. It is so called from its discov- 
erer, Dr. Brook Taylor, and was first published by him in 
1715, in his Method of Increments. 

Hence, by Taylor's Theorem, we may develop a function 
of the sum of two variables, as u' =f(x + y), into a series 
of terms, the first of which is the value of the f auction 
when y — ; the second is the value of the first derivative 
of the function when y = 0, into y ; the third is the value 

of the second derivative when y = 0, into ----, etc. 

The development of f(x — y) is obtained from (6) or (7), 
by changing + y into — y; thus, 



BINOMIAL THEOREM. 97 

day tV-u y 2 d 3 u y 3 

or, f(x-y)= f(x) -f (x) \ +f» (x) f- _/'" (x) -J^ 

+ ^^1^374-^ 
Ooe. — If we make x = in (8), we have 

«'=/(*)=/(0)+/'(0)f+/"(0)^-+/"'(0) r ^-3 

which is Maclaurin's Theorem. See (7) of Art. 58. 

THE BINOMIAL THEOREM. 
67. To Develop %i* = (as + y) n . 

Making y = 0, and taking the successive derivatives, we 

have 

f(x) = x n , 

f (x) = fMf- 1 , 

/"(a) = »(» — l)-af-« 
/'"" (x) = n (n -l)(n- 2) x n ~ 3 , 
f (x) = » (« — 1) (w — 2) (n — 3) x n -*, 
etc. etc. 

Substituting these values in (7), Art. 66, we have 

u' = (x + y) n = x n + — -— 7 + -i — --^ 2- 

w (K - i) ( n -2)a"- 3 2/ 3 
+ 1T ^ 3 4- etc., 

which is the Binomial Theorem (see Art. 59). 
5 



98 APPLICATIONS OF TAYLOR'S THEOREM. 

68. To Develop u' — sin (as + y). 

Here / (x) = sin x, f (x) = cos x, 

f" (x) = — sin x, f" (x) = — cos x, etc. 

Hence, 

u' = sin (x + «/) 

= sin * ( x - ft + 1& - 1.2/4.5.6 + etc ) 

+ C ° S X (f - ifa + IT^fes - T^rfceT? + etC ) 
= sin# cosy + cos# sin?/. (See Art. 60.) 

THE LOGARITHMIC SERIES. 

69. To Develop u' = log (as 4- y). 

Here 



Hence, 



./» = log a, 




r(*) = |. 


/w=| 




/"<*) = --, 


/"(*) = -i> 




etc. 


«' = log (x 


+ y) 





= log * + ! - s 3 + -A - !s + etc - 



x 2x 2 3x 3 4 ar 



Coe. — If $=1, this series becomes 
log(l+2/)=f-| 2 + | 3 -| 4 + etc, 
which is the same as Art. 61. 



EXPONENTIAL SERIES. 

70. To Develop u' = a x+ s/. 

Here /(a;) = «*, f" (x) = a? log 2 a, 

/' (2) = a* log a, f" (x) = a* log 3 a, etc. 



FAILURE OF TAYLOR'S THEOREM. 99 

Hence, 

u' = a x+ v 

= ^(1 + loga-y + log 2 «^ + lo g 3 «f^3 + etc -)- 

Cor. — If x = 0, this series becomes 

av — 1 + log a-y + log 2 a~ + H 3 ^;^ + etc., 

which is the same as Art 62. 

71. Though Taylor's Theorem in general gives the cor- 
rect development of every function of the sum of two 
variables, yet it sometimes happens that, for particular 
values of one of the variables, the function or some of its 
derivatives become infinite ; for these particular values, the 
theorem fails to give a correct development. 



For example, take the function u' = Va -f x + y* 
Here, f{x) = Va -f x, 



/" (*) = - 



%Va + x 

1 



4 (a + 3)*' 



/'"(4= ~— -T 5 -> etc. 
8 (a + x)* 

Substituting in (8) of Art. 66, we have 



u' = <\/a + x + y 

= V^T^ + —7^= £—g 4- ^— - etc. 

2V« + £ 8(a + z)* 16(6? + ^)2 

Now when a; has the particular value — a, this equation 
becomes 

u' = \fy = -f 00 — Qo + 00 — etc. ; 



100 EXAMPLES. 

that is, when x = — a, Vy = <*> . But y is, in dependent of 
x, and may have any value whatever, iife^pective of the 
value of x, and hence the conclusl&frjthat when x = — a, 
V y — oo , cannot he true. Foj .every other value of x, 
however, all the terms in the series will be finite, and the 
development true. 

Similarly, u' == a -f \/a — x -\- y gives, when substituted 
in Taylor's Theorem, 



u' = a + Va — x+y 

= a + Vet — x ^ - + etc., 

V a — x 

which, when x = a, becomes 

u' = a + Vy = a — oo -f etc. ; 

and hence the development fails for the particular value, 
x = a. 

It will be seen that when Taylor's Theorem fails to give 
the true development of a function, the failure is only for 
particular values of the variable, all other values of both 
variables giving a true development ; but when Maclaurin's 
Theorem fails to develop a function for one value of the 
variable, it fails for every other value. 

Many other formulas, still more comprehensive than these, 
have been derived, for the development of functions; but a 
discussion of them would be out of place in this work. 

EX A M PLE S. 

1. Develop y = a/I + as 2 . 

Put x 2 = z, and develop ; then replace z by its value. 

Ans. y = Vl + %* 

x 2 x i x 6 5x 8 . , 

= 1 + 2 --4- + T6-m + etc - 



3 - » = 1_, 



EXAMPLES. 101 

1 

— l_|_ a ;J_ a ;2 + a: 3 + a 4_j_ e t C . 



3. y = (a f a:) -3 . 

y = (a + z)" 3 = a -3 — 3a _4 a; + 6a~ 5 x 2 — lOar -6 ^ 

+ etc. 
L y = e siax . 

x 2 x* x 5 x 6 



2 2-4 ' 3-5 2.4-5.6 
+ etc. 

5. y = ae*. 

y = ze* = a; + a; 2 + g- + ^ 3 + etc - 

6. y = V2a; — 1. 

# = \/2a; — 1 = V— l(l— £ — -— - — etc.). 

1. y= (a 2 + a*)i 

y = (a 8 4- a: 2 ) 3 = flT -j_ |«3^2 _j_ | ft -3 ^ 4 — ■gear's afi 

+ etc. 



V« 4 + ic 4 



A 1 1 a 4 5a- 8 5-9.T 12 

(«* + *)-* = --3-5 + 



4« 5 ' 4-8r. 9 4-8-12rt 13 

5- 9- 13a: 16 

etc. 



1 4-8.12.16fl 17 
9. y = (a 5 + aiv — x 5 )^. 

Put a*x — re 5 = 2, as in Ex. 1. 

4-9 X s 



9.14 ^ 

+ etc. 



r 5 3 a 2 1-2-3 
4-9-14 z 4 



5% 3 1-2-3.4 



1 02 EXAMPLES. 

10. u = (x + y)h 

u = x* -|_ \%~%y — ix~iy 2 -f -^art^s _ e £ c> 

11. u = cos (a + y). (See Art. 68.) 

u = cos (# -f- y) = cos sc cos y — sin a; sin y. 

12. «/ = tan x. 

x 3 2x 5 
y = tanx = x + ~ + — + etc. 

13. y = sec a;. 

_, a; 2 5a 4 61a; 6 
2/ = S ec^l+-+- + — + etc 

14. ^ = log (1 + sin x). 

X 2 X 3 X 

y = log (1 + sin x) = x — - + ^ - -^ + etc. 



CHAPTER VI. 



EVALUATION OF INDETERMINATE FORMS. 

72. Indeterminate Forms. — When an algebraic ex- 
pression is in the form of a fraction, each of whose terms is 
variable, it sometimes happens that, for a particular value 
of the independent variable, the expression becomes inde- 
terminate ; thus, if a certain value a when substituted for 

x makes both terms of the fraction ^-~4 vanish, then it 

o (x) 

reduces to the form - , and its value is said to be indetermi- 
nate. 

Similarly, the fraction becomes indeterminate if its terms 
both become infinite for a particular value of x; also the 
forms co x and go — co , as well as certain others whose 
logarithms assume the form co x 0, are indeterminate forms. 
It is the object of this chapter to show how the true value 
of such expressions is to be found. By its true value is 
meant the limiting value winch the fraction assumes when 
x differs by an infinitesimal from the particular value which 
makes the expression indeterminate. It is evident (Arts. 9, 
43) that though the terms of the fraction may be infinitesi- 
mal, the ratio of the terms may have any value whatever. 

In many cases, the true values of indeterminate forms 

can be best found by ordinary algebraic and trigonometric 

processes. 

x 2, — 1 
For example, suppose we have to evaluate - 7i when 

x = 1. This fraction assumes the form - when x = 1 ; 
but if we divide the numerator and denominator by x — 1 



104 EXAMPLES. 

\ 

before making x = 1, the fraction becomes - — ; 

and now if we make x = 1, the fraction becomes 

1+1+1 _3 
1 + 1 ~ 2' 

which is its true value when x = 1. 

73. Heuce the first step towards the evaluation of such 
expressions is to detect, if possible, the factors common to 
both terms of the fraction, and to divide them out ; and 
then to evaluate the resulting fraction by giving to the 
variable the assigned value. 

EXAMPLES. 

1. Evaluate 3? _^~ + \ x _ V when * = J- 

This fraction may be written 

(x — l)(a? + a; + l) x 2 + x + 1 Q . ■ 

■5 .() n —tt = -7T- ■ — t = 3, when x = 1. 

(x — 1) (x 2 — ic + 1) z 2 — a; + 1 

x 

2. The fraction — = -—==-. = -^ , when x = 0. 

V « + ■# — v « — •£ u 
To find its true value, multiply both terms of the fraction 
by the complementary surd, Va + x + V# — x, and it 
becomes 

x (Va + a + a/« — a) V«+jc + Va — a; 

Yx : or ~~ 2 " ■ ■ ; 

and now making x = 0, the fraction becomes V«, which is 
its true value when x == 0. 



2# — a/5^ 2 — a 2 , ^1 

3. — -, when x = a. Ans. -|~. 

a; — V%x 2 — a 2 



METHOD OF EVALUATION. 


V 




Ans. 5. 


~ , when x = l. 

x — 1 



6. V% 2 + ax — x, when x = oo ^4ws. x- 

There are many indeterminate forms in which it is either 
impossible to detect the factor common to both terms, or 
else the process is very laborious, and hence the necessity of 
some general method for evaluating indeterminate forms. 
Such a method is furnished us by the Differential Calculus, 
which we now proceed to explain. 



METHOD OF THE DIFFERENTIAL CAL- 
CULUS. 

74. To evaluate Functions of the form =• 

Let f(x) and (f> (x) be two functions of x such that 
f(x) = and (x) = 0, when x = a. 

Then we shall have ^Vr = k' 
<p(a) 

Let x take an increment 7i, becoming x + h ; then the 
fraction becomes 

f(x + h) 
♦ (« + *)' 

Now develop /(a? + A) and (sc + 7i) by Taylor's Theo- 
rem ; substituting h for y in (7) of Art. 66, we have 

♦(».+ *> *(*) + * <*)*+>(*>£ + eta' 



106 METHOD OF EVALUATION. 

or when x = a, 

/(g + A ) _ /W +/ 1 W »+/"(«)? + t°- (i) 

(<? + A) (a) + 0' (a) 7* + 0" (a) * + etc. 

But by hypothesis / (a) = 0, and (a) = 0. Hence, 
dropping the first term in the numerator and denominator, 
and dividing both by A, we have, 

/( a + A)^ / ' (a)+/ " (a) ^ + etc - • 

Now when 7i = 0, the numerator and denominator of 
the second member become /' (a) and </>' (a) respectively ; 
hence we have, 

f(a) f(a) 

(a) </>' («)' 

as the true value of the fraction •* , ..\ , when x = a. 

(1.) If f (a) = and 0' (a) be not 0, the true value of 

^— 7-t is zero. 
*■(*) 

(2.) If/' («) be not zero and 0' («) = 0, the true value of 

f(a) . 

'!-f-{ IS GO. 

(3.) If /' {a) = 0, and 0' («) = 0, the new fraction 

4— r is still of the indeterminate form -• Dropping in 
(«) 

this case the first two terms of the numerator and denomi- 
nator of 
we have, 



u \ \ is still of the indeterminate form -• Dropping in 

e n 

W 
nator of (1), dividing both by jr 9 and making h = 0, 



EXAMPLES. 107 

f(a) = /»fr) 

0(«) </>'»' 

as.ihe true value of the fraction Vt-t j when x = a, 

4>(x)' 

If this fraction be also of the form - , we proceed to the 

next derivative, and thus we proceed till a pair of deriva- 
tives is found which do not both reduce to zero, when 
x s= a. The last result is the true value of the fraction. 



EXAMPLES. 

1. Evaluate — — - , when x = 1. 

x — 1 

Here / (x) = log x, </> (x) = x — 1 ; 

/. /W=p and ^(a;) = l; 



= 1. 

i* 



1 

(#) </>' (#) ~~ 1 £ 

That is, — 2__^ = 1, when x = 1. 
x — 1 



2. Evaluate — L , when x = 0. 



/"' (a) sin a; n 

VH = ~^r~ = a) wnen x = 0; 
(#) 2# 7 

/" (aj) __ cos x 



= i- 



Therefore, £^~| : 

0(£)Jo 



* The subscript denotes the value of the independent variable for which the 
function is evaluated. 



108 



EVALUATION. 



' . TT 

x sin x — - 

3. Evaluate , when x = T . 

cos a 2 



Here 



/' ( x ) _ x cos x + sm ^ 
</>' (2;) ~ — sin x 



-J2 



* — 1 






Hence ^ = - 1. 



' a * — 5» ' 
4. -, when # = 0. 



-4^5. log t- 



5. 7 r:, when x = a. 



(x — of' 



Here 



ne nx 



0' (x) s (x — a) s . 



— 1- 



00 or 0, 



according as s > or < 1. 






x — sm sc , 
6. , when x = 0. 



8. 



# — sin # 

e x — 6 



, when x = 0. 



sm # 



, when sc = 0. 



-4w5. 2. 
Ans. 2. 



Yq ^ZLl 



sma; 



x — sm x 



, when x = 0. Take the third 

Jbzs. 4. 



r ^derivative. 

^-J 10. ^ ^-, when a; = a. Cancel the factor (a — x)t 



(a — x)^ 



Ans. (2a)i 



75. To evaluate Functions of the form |^. 



Let 



f(x) oo 

T-7-T = — > when # = a. 

(p (x) 00 



E VAL UA TION. 109 

Since the terms of this fraction are infinites when x = a, 
their reciprocals are infinitesimals (Art. 8) ; that is, 

= 0, and — 7-7- = 0, when x = a ; hence, 



fix) " ' 0(*) 

1 






(x) 
and therefore the true values of ^ ; may be obtained by 

7¥) 

Art. 74 ; that is, by taking the derivatives of the terms, 
thus, 

1 <P'(x) 

/(*) _*W_ [»(s)]» _ 0' (*)[,/»? , _ 
*(*) - _l - /' (») -/-(*) [*W wnen ~ a ' 



or 



«(«) _ /'(«)[*»(«)] 3 



Dividing by • , "we get, 
J (a) ° 

"/'(a) ' 0(a)' 

whenoe /Ol) _/>). 

0(«)-0'(a) 

Hence the true value of the indeterminate form — is 

00 

found in the same manner as that of the form - • 

In the above demonstration, in dividing the equation by jVt, when 

x = a, we assumed that —— is neither nor 00 . so that the proof 

9(«) r 

would fail in either of these cases. . ,. — 



110 EXAMPLES. 

It may, however, be completed as follows : Suppose the true value 

of ^-7^ to be ; then the value of ^-^ — — - — - is 7i, where h may 

9 («) <P {a) 

be any constant. But as this latter fraction has a value which is 

neither nor go, its value by the above method is — - — —, s 

<p{a) . 

=V7- — h h; and since the value of this fraction is h, the first term 
9 '{a) 

^=--0;i. c, where M = , f^S is also 0. 

<p (a) 9 (a) <b f {a) 

fix) 
Similarly, if the true value of ^- be co when x = a, then 

9(a) 

— - - = 0; and therefore we have / x = 0, by what has just been 
J\ a ) J \ a ) 

' ' i (a) 

Therefore, in every case the value of ^-ryhr determines the value of 

9 (a) 

~-{ for either of the indeterminate forms ' K or — . (See Williamson's 
9 («) 00 v 

Dif. CaL, p. 100.) 

EXAMPLES. 
lo2f X 

1. Evaluate — ^— , when x = 00 . 

X' 1 



Here ^ = ■»" = -*-, = : -!~| « a 

(&•) (#) rase* -1 nx n _j a) 

log a; 



102" $/ 

2. Evaluate — ^ — , when x = 0. 
cot x 



Here 



/'(*) 


£ 


sin 2 a? 








0' (x) - 

f" (*) 


— cosec 2 ^ ~~ 
2 sin a; cos x~ 


— • 


0' 


*" (*) ~ 


1 








\ogx _ when ^ _ a 
cot X 





EVALUATION. 


111 


3. 


1 — log X , 

r 5 — > when x = 0. 

gx 


Ans. 0. 


4. 


7T 

, when x = 0. 

cot y 


A ** 



log tan (2#) , 
5. , 5 , v y , when x — 0. ^*. 1. 

log tan & 

76. To evaluate Functions of the form x co . 

Let f{x) x <f>(z) = x oo, when x = a. 
The function in this case is easily reducible to the form 
- ; for if /(«) = 0, and (a) = oo , the expression can be 

written ^i~i, which == -; therefore fix) x $(x) = ^\^, 

(fl) 0(F) 

may be evaluated by the method of Art. 74. 

EXAMPLES. 

TTX 

1. Evaluate (1 — x) tan — , when x = 1. 

We may write this 

1- a 

. fix 
cot^ 



_ 2 



„ /'(*) -l 2 sin2 (f ) 
Here :z#/\ = = 

(X) TT TTX TT 

v — 5 cosec 2 -^r- 

2. Evaluate se n log #, when # = 0. 

1 

„ , log X~\ X x n ~~\ 

X n log X = -5- = —r = =0. 



112 EXAMPLES. 

3. e~* log x, when x = oo . Ans. 0. 

4. sec x Ix sin x — ^J, when a; =.-• ^Iws. — 1. 

77. To evaluate Functions of the form co — oo. 

Let f(x) and <j> (x) be two functions of x which become 
infinite when x = a. Then f(x) — (#) = oo — oo , when 
x = a. 

The function in this case can be easily reduced to the 

form - , and may be evaluated as heretofore. 

EXAMPLES. 

2 1 

1. Evaluate -= . when x = 1. 

x 2 — 1 x — 1 

This takes the form oo — oo , when x = 1. 

2 1 2— a— 1 

, when « = !. 



£2_1 z_i x 2 —l 

T7-/— r = -it- = — h when a? = 1. 

</>' (a) 2z 2 ' 

^ -r^ ! x 1 ' lOff (1 + Z) 

2. Evaluate .; , — r ^—5 , when x = 0, 

sc (1 + x) x 2 

which takes the form 00 — 00 , when x = 0. 

1 _ _ l°g (1 + x ) _ x ~ 0- + ») log (1 + x) 
x (1 + x) x 2 x 2 (1 4- x) 

_ a; — (1 + a?)log(l + x) 

~ x 2 

(remembering that 1 + # = 1, when x vanishes). 





EVALUATION. 113 


<p'(x) 


_._l-log(l+z)-l_0 ^ J, 
~ 2x ~ 0' J 




* X 1 




<t>"{x) 


1 + a; 

~ 2 


o--*- 


Evaluat 


e sec # — tan a?, when a; = ^ • 


sec x - 


1 — sin x , 7r 

- tan x = ■ = - , when x = - • 

cos # 2 






- COS x~\ 




- sin a: L ~ 

~J 2 



7T 77 

Hence, sec s and tan - are either absolutely equal, or 

differ by a quantity which must be neglected in their alge- 
braic sum.* 





Aiis. \. 


x — 1 log X 


z. z , when x = 1. 

log x log X 


Ans. — 1 



78. To evaluate Functions of the forms 0°, co°, 
and l ±c0 . 

Let f(x) and <f>'(x) be two functions of x which, when 
x = a, assume such rallies that [/(a)]* (ar) is one of the 
above forms. 

Let y = [f(x)] Hx) ; 

.-. logy == (p(x)\ogf(x). 
(1.) When f(x) = oo or 0, and (x) = 0. 

log y = 9 (*) log/W = (± oo ), 
which is the form of Art. 76. 

* Price's Infinitesimal Calculus, Vol. I, p. 210. 



114 EXAMPLES. 

Hence, [/(#)] becomes indeterminate when it is of 
the form 0° or oo °. 

(2.) When f(x) = 1, and (x) = ± oo . 

log y = (f) (x) logf(x) = ± oo x 0. 
Hence, [/(»)]* is indeterminate when of the forms 

Hence the indeterminate forms of this class are 

0°,* a>°, 1 ±0 °, 

and may all be evaluated as in Art. 76, by first evaluating 
their logarithms, which take the form x oo . 

EXAMPLES. 

1. Evaluate x x , when x = 0. 

We have log x x = x log £ = _ t ; 

1 

/'(#)_ # l _ n . 

— 7—, r — = — — 3? Jo — U 1 

</> (a?) — ar 2 J 

.\ log x x = 0, when a; = 0; 
hence, x x == 1, when a? == 0. 

2. Evaluate a x , when a; = 00 . 

v 1 1 log SC 

log af = - log x = -|- ; 



1 

/>_) = x_ = I 
<p' (a) 1 " x_ 



= 0; 



* In general, the value of the indeterminate form 0° is 1. (See Note on Inde. 
terminate Exponential Forms, hy F. Franklin, in Vol. I, No. 4, of American Journal 
of Mathematics.) 



EXAMPLES. 115 



.*. log x* = 0,. when x = oo ; 
hence, of = 1, when x = oo . 



(> + gr. 



3. Evaluate (1 - ) , when x = oo . 

Let x = - . and denote the function by u. 

z 

Then u = (1 '+ azf\ 

(since when x = oo , 2 = 0) ; and 

, log (1 + az) , A 

log w = — — — ! £ , when z = 0. 

Taking derivatives, we have 
a 



log u c 



= a; 

o 



1 + «2_ 

1 + - ) == a, when a; = oo ; 

(a\ x 
1 + - j = e a , when x = oo . 

If « = 1, we have 

that is, as a; increases indefinitely, the limiting value (Art. 
41) of the function ( 1 -f - 1 is the Naperian base. 

(-| \ tan x 
) , when x = 0. Ans. 1. 

5. af inx , when a; = 0. Ans. 1. 

o»\tan — £ 

2 _ ±\ 2a , when a; = QJ Ans. e\ 



116 COMPOUND INDETERMINATE FORMS. 

79. Compound Indeterminate Forms. — If an inde- 
terminate form be the product of two or more expressions, 
each of which becomes indeterminate for the same value of 
x, its true value can be found by evaluating each factor 
separately ; also, when the value of any indeterminate form 
is known, that of any power of it can be determined. 

EXAMPLES. 

x n 
1. Evaluate — , when x = co. 
e x 



(x\ ri 
~) 

X 

We first evaluate — , when x = oo 



en 



(x) 1 



n 



= 1=0. 

00 



GO 



Hence, -" = n = 0. 

2. Evaluate x m log" x, when x = 0, and m and n are 
positive. 

Here [x n log a)" = ( — ^~ J • 

\ aT» / 

Iop* #* 
We first evaluate — ^- , when x = 0. 

1 

We have ^4 " 



-= 0. 



x n 


n 


n -~ 
x n 




m 






.\ x m log 1 x — 0" = 0. 



EXAMPLES. 



117 



3. 



UjM sgn 



, when x = 1. 



This function can be written in the form 

x m I _ x n 



1 + X* l — XP 

We have to evaluate only the latter function for x = l, 
since the former is determinate. 



Here 



fM __ ~ Ms"" 1 - ? _v 



0'(3) 



— «$ p- J 



# 



a;" 1 _ 1 
l+^"2' 



a?" — x 1 ' 



x-p 



n 



when x = 1. 
when a; = 1. 
when sc = 1. 



(a^ — cr) sin t— 
. v y 2a 
4. — , when x — «. 

«, ^a; 

a; 2 cos 7T- 

2a 



(x 2 — a 2 ) sin 



nx 
2a 



. 7ra; 
a; 2 — a 2 2a 



X 2 cos 



7TX 

2a 



cos 



2« 



"We have only to evaluate the first factor, 



x 2 — a 2 ' 

TTX 
COS^r- 

2a _ 

nx~ 



2x 



7T . TTX 

2a 2a 



±a 2 



sin 



and 



2a 



118 EXAMPLES. 



(*-.*) ring ^ 

nx TT 

X 2 COS — 

2a 



EXAMPLES. 



log (1 + X) ' 



when a? = 0. ^4ws. 2. 



2. —5 § Q , when a; = 3. J. 

a; 3 — a; 2 — ox — 3 * 



/sin wajX" 1 



3. ( -1 , when x = 0. # m . 



4. — t— * (dif. three times), when a; = 0. 

h 

w 1 — sin x + cos a? , 7r 

5. — =■ , when «"=^« ^ 1. 

sm jp + cos a; — 1 2 

tan a; — sin a; 

o. — r , when a; = 0. *. 

sin 3 a; •* 

$ sm * — 05 7T . 

7. i : — , when x = - • a log #. 

log sm a; 2 ° 

Q £ 2 + 2 cos a; — 2 ,,.„ . ... x -, . 

8. (dif. four times), when x = 0. 

a- 

9. a; 1- * (pass to logarithms), when x = 1. 

1 

10. «**(!*«) , whena; = 0. 3. 

1 — COS X 

11. a;-e ¥ (pass to logarithms), when x = 0. 



EXAMPLES. 119 



12. ( I (pass to logarithms), when x = ao. 

Ans. 1. 

13. 2a: sin — , when a; = oo . a. 

14. e sin x, when a; = 0. oo . 

15. (cos ax) cosec2cx (pass to logarithms, and dif. twice), 
when x — 0. - — 

16. a?" (sin xY*« x ( jT^fY (see Art. 79), when a; = £• 

v y \2 sin 2a:/ v /3 2 

2'*-i-3* 

17. (sin a:) tan *, when & = £• 1. 

' m*f~J (~^> 

2. 



CHAPTER VII. 

FUNCTIONS OF TWO OR MORE VARIABLES, AND 
CHANGE OF THE INDEPENDENT VARIABLE. 

80. Partial Differentiation. — In the preceding chap- 
ters, we have considered only functions of one independent 
variable; such functions are furnished us in Analytic 
Geometry of Tivo Dimensions. In the present chapter, we 
are to consider functions of two or more variables. Analytic 
Geometry of Three Dimensions introduces us to functions of 
the latter kind. For example, the equation 

z = ax -f by + c (1) 

represents a plane ; x and y are tivo independent variables, 
of which z is a function. In this equation, z may be 
changed by changing either -x or y, or by changing them 
both, as they are entirely independent of each other, and 
either of them may be considered to change without affect- 
ing the other ; in this case z, the value of which depends 
upon the values of x and y, is called a function of the inde- 
pendent variables x and y. 

A ^partial differential of a function of several variables is 
a differential obtained on the hypothesis that only one of 
the variables changes. 

A total differential of a function of several variables is a 
differential obtained on the hypothesis that all the variables 
change. 

A partial derivative of a function of several variables is 
the ratio of a partial differential of the function to the dif- 
ferential of the variable supposed to. change. 



PARTIAL DIFFERENTIATION. 121 

A total derivative of a function of several variables is the 
ratio of the total differential of the function to the differen- 
tial of some one of its variables. (See Olney's Calculus, 
p. 45.) 

As all the variables except one are, for the time being, 
treated as constants, it follows that the partial differentials 
and derivatives of any expression can be obtained by the 
same rules as the differentials and derivatives in the case of 
a single variable. 

If we differentiate (1), first with respect to x. regarding 
y as constant, and then with respect to y, regarding x as 
constant, we get 

dz = adx, (2) 

and dz = My. (3) 

Dividing (2) and (3) by dx and dy respectively, we get, 

£ = "' (4 > 

and -J- == b. (5) 

dy ' 

The expressions in (2) and (3) are called the partial 

differentials of z with respect to x and y, respectively, while 

d" dz 

-7 1 and - T - are called the partial derivatives of z with re- 

dx dy 

spect to the same variables. 

Since a and b in (4) and (5) are the partial derivatives of 
z with respect to x and y, respectively, we see from (2) 
that the partial differential of z with respect to x, is equal 
to the partial derivative of z with respect to x multiplied 
by dx, and similarly for the partial differential of y. 

Hence, generally, if 

/ i x > V> z) 

denotes a function of three variables, x, y, z, its derivative 
or differential when x alone is supposed to change, is called 
6 



122 PARTIAL DIFFERENTIATION. 

the partial derivative or differential of the function tvith 
respect to x, and similarly for the other variables, y and z. 
If the function be represented by u, its partial derivatives 
are denoted by 

du du du 

dx * dy ' dz ' 

and its partial differentials by 

du 7 du , du _ 

-7- fla;, -7- dy, -=- «& 
fl# «y J dz 

81. Differentiation of a Function of Two Varia- 
bles. — Let u = f (x, y), and represent the partial differ- 

du 
ential of u with respect to a?, by -y- fl#, and with respect 

din 
to y, by -r- fly, while o*m represents the total differential. 

Let x and y receive the infinitesimal increments dx and 
dy, and let the corresponding increment of u be du. Then 
we have, 

du =f(x + dx, y + fly) — /(«, y). 

Subtract and add / (x, y + fly), and we have 

oe« ^-/{x+dap, y + dy)—f{x, y + dy) + f(x, y + dy) 

-/(%> y)- 

du 
Now f(x + dx, y + dy)—f{x, y + dy) = -=-dx, because 

it is the difference between two consecutive states of the 
function- due to a change in x alone; that is, whatever dif- 
ference there is between f{x + dx, y + dy) and /(a;, y + dy) 
is due solely to the change in x, as y + dy is the value of y 
in both of them. For the same reason, 

du 

f{x, y + dy) -f(x,y) = ^dy; 

and therefore we have du = -7- dx + -7- dy, 

dx ay 



EXAMPLES. 123 

in which -r- dx and ~ dy are the partial differentials of u 

with respect to x and y, respectively, while du is the total 
differential of u when both x and ?/ are supposed to vary. 
In the same way, we may find the differential of any num- 
ber of variables. 

Hence, the total differential of a function of two or more 
variables is equal to the sum of its partial differentials. 

The student will carefully observe the different meanings 
given to the infinitely small quantity du in this equation, 
otherwise the equation will seem to be inconsistent with the 

principles of algebra. Thus, in -=- dx, du denotes the in- 
finitely small change in u arising from the increment dx in 

du 
x, y being regarded as constant. Also, in -j- dy, du denotes 

ay 

the infinitely small change in u arising from the increment 
dy in y, x being regarded as constant, while du in the first 
member denotes the total change in u caused by both x and 
y changing. If the partial differentials of x and y be rep- 
resented by d x u and d y ii, respectively, the preceding equa- 
tion may be written 

du = d x u + d y u. 



EXAMPLES. 

1. Let u = ay 2 -f bxy + ex 2 + ey -f gx + k, to find the 
total differential of u. 

Differentiating with respect to x, we have 

d x ii = bydx + 2cxdx -f gdx. 
Differentiating with respect to y, we have 

d y u = 2aydy + bxdy + edy. 
Hence, du = (by -f 2cx + g) dx + (2ay + bx -f c) dy. 



124 EXAMPLES. 

2. U = X*. 

Here d x u = yx y ~ ] dx, d (J u = xv logx dy. 
Hence, du = yx y ~ l dx + x y log x dy. 



3. u = 


'x 2 y 2 
a 2 + ¥ 


Here 


d x u = -^dx, dM = -£dy. 
a 2 b 2 


Hence, 


du = —zdx + ^r dy. 
a 2 o 2 J 


4. u = 


: tan" 1 £ 

X 


Here 


ydx 

x 2 ydx 


d * U - y2~ x 2 + y 2 ' 
1 + 2 




% 




^ x xdy 




L ^ X 2 


Hence, 


^ xdy — ydx 
x 2 + y 2 



, x ■ i y 

5. u =. sin -1 - 4- sin -1 %• 
a o 

Here d x ii = — - , d,,u = — - — — 

Va« - x 2 V6 2 - 

TT 7 *?# dy 

Hence, du = 



V# 2 — # 2 v^ 2 — # 2 

, ., sin x dy 

6. w = 2/ s,nx . du = y imx logy cos x dx -\ — « 

* ^ n * ^/ covers x 



TOTAL DERIVATIVE. 125 



<x , udx — xdit 

7. u = vers -1 -■ du = • m - z — 

8. u = log ■&. du = - dx + log z %. 



82. To Find the Total Derivative of u with re- 
spect to x, when u = f (y 9 &), and y = <p (x), 
z = fa (ac). 

Since w =f(y, z), we have (Art. 81), 
, du .. du -, 

dw = Ty dy + Tz dz > W 

and since «/ = </>(#), we have e?# == -~dx\ 

since z = fa (x), we have dz = ~ dx. 

(XX 

Substituting these values for dy and *7z in (1), we get 

n du du , du dz 7 

die = — -f- dx -f -=- ^- - dx. (2) 

dy dx dz dx ' 

Dividing by dx, and denoting the total derivative by ( ), 
w T e have 

(du\ _ du dy y du dz 
dx) ~ dy dx n dz dx ^ ' 

Cor. 1. — If z — x, the proposition becomes u =f(x, y) 

dz 
and y = <p (x) ; and since y = 1, (3) becomes 



(du\ __ du du dy 
dx) ~~ dx dy dx 



dy 

Cor. 2. — If u =f(x, y, z), and y = <p (x), and z = fa (x), 
we have 

7 du 7 du 1 du 7 ,„* 



126 EXPLANATION OF TERMS. 

dy = — - dx, and dz = -=- dx. 

dx dx 

Substituting the values of dy and dz in (1), and dividing 
by dx, we get 

(du\ __ du du dy du dz 

\dx! dx dy dx dz dx 



Cor. 3.— If u = f(y, z, v), and y = (j> (x), and z = 
0i (x), and t; = S (a), we have, 

7 df^ , du 7 , du -, 
du = Ty dy + Tz dz + Jv dv. (1) 

, dy , , 6?« 7 7 £fo> 

dy = -y- dx ; «2 = -y- ## «# = -r- dx. 

* dx dx dx 

Substituting the values of dy, dz, dv, in (1), and dividing 
by dx, we get 

idu\ __ du dy du dz du do 
\dxl dy dx dz dx dv dx 

du 
Cor. 4. — If u = f(y) and y = (x), to find -r-« 

dx 

Since u = f(y), we have du = -7- c?y. 

a?/ 

Since y = (a;), we have dy = -^ dx. 

therefore, du = - T --y- dx, and .*. -7- = -r-i 

dy dx dx dy dx 

Sch. — The student must observe carefully the meanings 
of the terms in this Art. Thus, in the Proposition, u is 
indirectly a function of x through y and z. In Cor. 1, w is 
directly a function of x and indirectly a function of x 
through y. In Cor. 2, & is directly a function of x and 



EXAMPLES. 127 

indirectly a function of x through y and z. In Cor. 3, u is 

indirectly a function of a; through ?/, 2, and v. In Cor. 4, 

w is indirectly a function of x through y. 

The equations in this Article may seem to be inconsistent 

with the principles of Algebra, and even absurd ; but a little 

reflection will remove the difficulty. The du's must be 

carefully distinguished from each other. In Cor. 1, for 

dn 
example, the die in -y- is that part of the change in u 

which results directly from a change in x, while y remains 

dii 
constant ; and the du in — is that part of the change in u 

which results indirectly from a change in x through y ; and 

(du\ 
-T-f is the entire change in u which results 

directly from a change in x, and indirectly from a change 
in x through y. 



EXAMPLES 
X 



and y = (r 2 - x^, to find (^V 



1. u = tan" 1 

Here £ = £, £=-^£ and 4 = _ * 

a# r 2 6/?/ r 2 ' ^ # 

^Substituttag in (|) = | + J | (Art. 82, Co, 1), we 

©-M-5K-3 

_ */ 2 + £ 2 1 



r 2 y Vr*-Fx*' 

and this value is of course the same that we would obtain 



x 



if we substituted in u — tan" 1 - for y its value in terms of 

y 

x, and then differentiated with respect to x. 



128 EXAMPLES. 



2. u=z tan -1 (xy) and y — e x , to find l-j-)' 

Here — = y — — = ?-- S-=* 

tto 1 -f a% 2? <% 1 + #y d# 



(Art. 82, Cor. 1), 



idu\ _ y 
\dxl ~ 1 



% 2 ~~ 1 +xW*' 



and this value is of course the same that we would obtain if 
we differentiated tan -1 (xe x ) with respect to x. 

3. u = z 2 + y s + zy and z = sin x, y — e x , to find (-t-)* 

it du n , du ' , 

Here ^ = V + * * = ^ + y, 

dz dy 

— = cos a-, -^ = e*. 

.-. (Art. 82), 

— (3 e 2a? _|_ s i n #) e a; _|_ (2 sin X + 0®) COS X 

= S&te -f e® (sin a -f cos 2;) + sin 2x. 
(See Todhunter's Dif. Cal., p. 150.) 

Let the student confirm this result by substituting in u, 
for y and z, their values in terms of x, thus obtaining 

u — e sx _j_ e a? s j n /p _j_ gin 2 a;, 
and then differentiate with respect to x. 

4. w = sin -1 (y — z), y = 3x, z = 4a; 3 . 

du __ 1 

dy ~~ ^/i-zr^rz."^) ' 



EXAMPLES. 




du 1 




az Vi — (# — s)' 




^ - 3 — = 12z 2 . 
dx ' dx 




'du\ _ 3 — 12a: 2 




dx) yi _ (y _ s) 




3 - 12a; 2 





129 



3 



VI — 9a: 2 + 24a 4 — 1 6a* Vl 



e ra (2/ — z) 
« 2 -fT 



„ e — vu — 6i 

5. W = 2T~i anc * # ~ a sm ^ ^ — C0S X ' 



dy a 2 + l' dz a 2 + 1' 



dz 
dx 

die ae ax (y — z) 



-f- = a cos x, -*- •= — sm a\ 

«a; «# 



dsc # 2 -h 1 

.-. (Art. 82, Cor. 2), 

-r = -= - (a cos x + sm # -f « 2 sm x — « cos x) 

\dx! a 2 + 1 v ; 

= e** sin x. (See Courtenay's Cal., p. 73.) 

6. w = ^ and y = e x , z — z A — ±3? + l&e 2 — 24a; + 24. 

© = ** 

7. If u —f(z) and 2 = (#, ?/), show that 

du dz 7 dfo dz , 
az fife Gfe «?/ * 



aW a 4 ?/ ar 4 
8. a = -^ - -± + — and y = log a;. 

6) = ^ (log x) 



130" PARTIAL DIFFERENTIATION. 

83. Successive Partial Differentiation of Func- 
tions of Two Independent Variables. 

Let u = f{x, y) 

be a function of the independent variables x and y ; then 

iiii (Jit 

-j- and — are, in general, functions of x and y, and hence 

ctx ay 

may be differentiated with respect to either x or y, thus 
obtaining a class of second partial differentials. Since the 
partial differentials of u with respect to x and y have been 

Ult (J 11 

represented by -j- dx and — dy (Art. 81), we may repre- 
sent the successive partial differentials as follows : 

The partial differential of (-r- dx), with respect to x, 

-—(—dx)dx 
which may be abbreviated into 

— dx 2 

dx* aX ' 

The partial differential of (— dx), with respect to y, 
d idu , \ , 

which may be abbreviated into 

dho , , 

~ n — j- ay dx. 
dy dx J 

/72'W d^U 

Again, both y-g dx 2 and - — — dy dx will generally be 

functions of both x and y, and may be differentiated with 
respect to x or y, giving us third partial differentials, and 
so on. Hence we use such symbols as 



PA B TIA L DIFFERENTIA TIOX. 1 3 1 

dy dx 2 , -j—j — r dx dy dx, and y , t % 2 e&, 



c??/ ffo 2 ^ tfa; % dz ' ' d# 2 eta' 

the meaning of which is evident from the preceding re- 

d u ' 

marks. For example, -j — ^ — -,- dx dy dx denotes that the 

function u is first differentiated with respect to x, supposing 
y constant; the resulting function is then differentiated 
with respect to y, supposing x constant; this last result is 
then differentiated with respect to x. supposing y constant; 
and similarly in all other cases. 

When u =f(x, y), the partial derivatives are denoted by 
d 2 u d 2 u d 2 u d 2 u d?n cPu 



dx 2 ' dy 2 ' dxdy' dx*' dx 2 dy' dxdy 2 ' 



etc. 



84. If u be a Function of x and y, to prove that 

cPu , , d 2 u 7 7 

^ — =- dx dy = -= — 7- dy dx. 
dx dy a dy dx a 

Take u = vfop, 

— dx = 2xy 3 dx, 

~du = ox 2 y 2 dy, 

d 2 u 

dy dx = 6xy 2 dy dx, 



dy dx 



dx dy = 6xy 2 dx 



dx dy 
In this particular case, 

d 2 u , _ (Pu _ _ 

- 7 — =- dx dy = - — =- dy dx ; 
dx dy J dy dx * 

that is, the values of the partial differentials are independ- 
ent of the order in which the variables are supposed to 
change. 



132 PARTIAL DIFFERENTIATION, 

To show this generally: 

Let u = f{x, y); 

then ~ dx = f(x + dx, y) —f{x, y). 

This expression being regarded as a function of y, let y 
become y + dy, x remaining constant ; then 

7i7,Y^ dx ) d y = f{% + dx, y + dy)— f{x, y + dy) 

= f(x + dx, y + dy) —f(x, y + dy) 

—f(x + dx,y) + f(x,y). 
In like manner, 

d$ dy ' = f( x > yJrdy "> ~f^ y)- 

TA^Jy)^ = f(* + dx > V + dy) -f(x + dx, y) 

V -;[/>, y + dy) -f(x,y)] 

— f(x + dx, y + dy) —f{x + dx, y) 

-f{x,y+dy)+f(%,y% 

These two results being identical, we have 



d (du , \ 7 d (du 7 \ 7 



dy\dx I J dx\dy 

that is, -= — 7 — dy dx = -t~i- dx dy, 

dy dx J dxdy ° 

Dividing by dy dx, we get 

d?u dhi 

dy dx ~ dx dy 

In the same manner, it may be shown that 
dhi 7 „ 7 dhi 7 7 9 

Wdy dx ^ = dfd^ y ' 
dhi d 5 u 

dx 2 dy dy dx*' 
and so on to any extent. 



EXAMPLES. 133 



EXAMPLES. 

1. Given u = sin (x + y), to find the successive partial 
derivatives with respect to x. 

du , , \ dhi ■ / , \ 

to = cos ( x + y)> -fai = - sm ( x + y)> 

dhi . N dhi . , * 

_ = _ cos (a; + y), -^ = sm (a + y), etc. 

1. w = log {x -\- y), to find the successive partial de- 
rivatives with respect to x, and also with respect to y in 
the common system. 

da m dhi m d?u 2m 



dx x + y' dx* (z f f/) 2 ' dx 3 (x + yf 

du _ m dhi _ m dhi _ 2m 

dy ~ x + y y dif ~ "~ (x + yf 3 d v* ~ ( x + yf 

(See Art. 65, Lemma.) 



dhi cPu 

dydx ~~ dxdy 

dhi cftu 

KijP *""v ""*-" dyhlx ~ dxdxf 

5. If u = sin (ax n + by 71 ), 



3. If u = x log y, verify that 

4. If u = tan -1 1-1, verify that 



verify that 



dhi dhi 



dx l dif dy 2 dx 2 



85. Successive Differentials of a Function of Two 
Independent Variables. 

Let u =f (x, y). 

We have already found the first differential (Art 81), 

7 du , , du . .... 

du ^^ dx + r s d ^ (1) 



134 SUCCESSIVE DIFFERENTIALS. 

Differentiating this equation, and observing that — ' — , 

dx dy 

are, in general, functions of both x and y (Art. 83), and 

remembering that x and y are independent, and hence 

that dx and dy are constant, we have, 

m d (P x dx ) 7 4S & ) 7 *($**), 

+- ^r* 

, 2 d 2 u 7 „ d 2 u , , <#% , 7 d 2 ^ , 9 

* M = a? ** + ^& **» + Txdy **» + ** <^ 

d 2 u d 2 u 

(since tt-t- t?y^ = -=— =- <£ec?#, Art. 84). (2) 

Differentiating (2), remembering that each term is a 
function of x and y, and hence that the total differential 
of each term is equal to the sum of its partial differentials, 
we get, 

and so on. It will be observed that the coefficients and 
exponents in the different terms of these differentials are 
the same as those in the corresponding powers of a bino- 
mial; and hence any required differential may be written 
out. 



* The total differential of each of the terms (-^ dx) and (~ dy\ is equal to the 
gum of its partial differentials, 



IMPLICIT FRACTIONS 135 





EXAMPLES. 




1. u = 


(x 2 -f- ?/ 2 )i 






du 


X 


du 

~dy ~ 


y 


dx 


' (r> + r)i' 


(x 2 + ^)i 


dht 
da? 




dhi 
dxdy 


— ay 

(* 2 + y 2 )? 


dhi 


<* 


dhi 
dtf ~ 


- Sxy 2 


df 


" c^ + ^r 


(x 2 + y 2 )i 


dhi 
dx 2 dy 


_ y {%& — y 2 . 

" (x 2 + ^)i ' 


dhi 
dxdy 2 ~ 


x (2y 2 — x 2 ) 
(* + */ 2 )* 


d 3 u 


- Syx 2 






dif 


" (a* + y 2 )* 





#>„ _ [_ Sxy 2 dx s + 3?/ (2a; 2 — y 2 ) dx 2 dy 
+ ox (2 if — x 2 ) dxdy 2 — 3yx 2 dy*] 



{x 2 + y 2 f 
2. u — e (ax ^ by) . 

dhi — [a 2 dx 2 + Zabdxdy + b 2 dif\ a*™* 

= [adx + bdy] 2 e ax4&i '. 

86. Implicit Functions (see Art. 6). — Thus far in this 
Chapter, the methods which we have given, although often 
convenient, are not absolutely necessary, as in every case by 
making the proper substitutions we may obtain an explicit 
function of x, and differentiate it by the rules in Chapter II. 
But the case of implicit functions which we are now to 
consider is one in which a new method is often indis- 
pensable. 

Let / (x, y) = be an implicit function of two varia- 

dii 
bles, in which it is required to find -^-- If this equation 



136 IMPLICIT FUNCTIONS. 

can be solved with respect to y, giving for example 
y = $ (x), then the derivative of y with respect to x can be 
found by previous rules. But as it is often difficult and 
sometimes impossible to solve the given equation, it is 

necessary to investigate a rule for finding -p without 
solving the equation. 

87. Differentiation of an Implicit Function. 

Let f(x, y) = 0, 

in which y is an implicit function of x, to find -p.- 

Let fix, y) — u. 

Then u ==/(«, y) = 0. 

Hence by (Art. 82, Cor. 1), we have, 

ldu\ __ du du dy 
\dxl ~~ dx dy dx 

But u is always = 0, and therefore its total differential 
du> 



= ; hence (-=- 1 = 0, and therefore, 
from which we get, 



du du dy _ 
dx dy dx ~ 



du 

dy _ dx 

dx ~ du 

dy 

Sch.— It will be observed that while y~\ = 0, neither 
-j- nor y- is in general = 0. For example, 

#2 + y 2 _ r 2 _ 

is of the form f(x, y) = 0. We see that if x changes while 
y remains constant, the function changes, and hence is no 



EXAMPLES. 137 

longer = 0. Also, if y changes while x remains constant, 
the function does not remain = 0. But if when x changes 
y takes a corresponding change by virtue of its dependence 
on x, the function remains = 0. 

EXAMPLTES. 

1, if — 2xy + a 2 = 0, to find ^ 



therefore, 



du 

dy _ dx — 2y y 

dx ~ du ~ 2y — 2x ~~ y — x 
dy 



2. a 2 y 2 + b 2 x 2 - a 2 b 2 = 0, to find ^- 

du 079 du 2 

Tx = Wx > * = **' 

du 

" « ffo/ ffo 2& 2 £ Z> 2 # ,_, , 

therefore, -/- = =-- = — ^— ,- = r - (1) 

aa; aw 2a 2 # a 2 y v 



Since ?/ = - Va 2 — x 2 , from the given equation, we may 
solve this example directly by previous methods, and obtain 

c ]! = _ hx (2) 

dx a^/ a 2_ x 2 } 

which agrees with (1) by substituting in it the value of y in 
terms of x. 

In this example we can verify our new rule by comparing 
the result with that obtained by previous rules. In more 
complex examples, such as the following one, we can find 

~ onlv by the new method. 
dx " J 



138 SECOND DERIVATIVE OF AN IMPLICIT FUNCTION. 

3. x 5 — ax z y + bx 2 y 2 — y 5 = 0, to find ■—- 
p. = 5x*-3ax 2 y + 2ta/ 2 ; 

,, „ dy 5x* — 3ax 2 y -f 2to 2 

therefore, -/ = -=-; — .. ^ 9 f- 

a# 5?/ 4 — 2ox 2 y -f «£ 3 

6??/ Sccx 2 A- 3x 2 'i/ 

5. # 2 - 2o»y + x 2 — I 2 = 0. $ = gy ~ g . 
^ ■ dx y — ax 



v. ^ — 


^ ' "^"' ^-3(1-^) 




7. *+**+,-* *_ J J*. 




88. To Find the Second Derivative of an 
plicit Function. 


Im- 


Let 
We have 




(1) 


or 


dte cfo/ dx ~" ' 


(2) 



it is required to find -~ 

Differentiating (2), remembering that -j- , -=- , are func- 
tions of a; and 2/, we get 

£?% d 2 ?* dy ld 2 u dy d 2 u \dy du d 2 y _ 
dx 2 dy dx dx \dy 2 dx dx dy) dx dy dx 2 



EXAMPLES. 139 

dhi _ f?% dy d 2 udu 2 du d 2 y _ /ox 

or -=-. + 2 - 7 — r JL + — -j- + — -2 = 0. (3) 

#2 2 a# dy dx dy 2 dx 2 dy dx 2 v ' 

Substituting the value of -=r from (1), and clearing of 
fractions, we get 

d 2 u du 2 dhi du du d 2 u du 2 du* d 2 y 

dx 2 dy 2 dx dy dx dy dy 2 dx 2 dy* dx 2 ~ 

Solving for -^, we get 

dhi /du\ 2 d 2 u du du d 2 u /du\ 2 

d 2 y _ dx 2 \dyj dx dy dx dy dy 2 \dx) 

dx 2 ~ 7dt(\* 

\dv) 



W 



dy) 

Sch. — This equation is so complicated that in practice it 
is generally more convenient to differentiate the value of the 
first derivative immediately than to substitute in (4). The 
third and higher derivatives may be obtained in a similar 
manner, but their forms are very complicated. 

Equation (2) is frequently called the first derived equation 
or the differential equation of the first order ; and equation 
(3) is called the second derived equation or the differential 
equation of the second order. 

EXAMPLES. 

1. y 2 - 2xy + a 2 = 0, to find ^ and ^. 

du du _ d?n . 

S = " 2 ^ Ty = 2 y- 2X '> dx~ 2 =°> 

dhi ' = -2; — =2. 



dx dy dy 

Therefore, by (1), % = -S- ; 

J v ' dx y — x y 

and bv m d2y - ~ 1Q y {y - x l±M - y d - 2x ) 

and by (4), -^ _ ^Z^i - (y - zf 



140 CHANGE OF INDEPENDENT VARIABLE. 

2. % f - 2axy + x 2 -¥ = 0, to find ^, |^ (See Sch.) 
dy ay — x 



d 2 y 



dx y — ax 

_ (y-«*)(«g-i)-(«y-*)(g -«) 

dx 2 (y — ax) 2 

_ (y — ax) (a 2 y — y) — {ay — x) (a 2 x — x) 
(y — ax) B 

(by substituting the value of ~) 

- (a 2 — 1) C.V 2 - 2«a?y + x 2 ) _ b 2 ( a 2 — 1) 
(y — «a;) 3 " (y — ax)' 3 

3. # 3 -}- 3axy -f ?/ 3 = 0, to find ~ and -~ 

ax ax 

Differentiating, we have 

(x 2 -f ay) dx + (y 2 -f «sc) dy = ; 

dy _ -a; 2 -f «# 

(2* + afj & + ax) - (gy J + «) (^+«y) 

flfo) 2 " (?/ 2 + tf^) 2 

= t-J^^-tv (See Price's Calculus, Vol. I, p. 142.) 

(y 2 + axf v 

4 * + «■_<*=<) dy = _x.d 2 y = _a 2 t 

u dx y ' 6^ 2 ?/ 3 

89. Change of the Independent Variable.— Thus 

dy dy 2 
far we have employed the derivatives -—,-—, etc., upon 

the hypothesis that x was the independent variable and y 
the function. But in the discussion of expressions contain- 



d 2 y 



VALUES OF DIFFERENTIAL COEFFICIENTS. 141 

ing the successive differentials and derivatives of a function 
with respect to x, it is frequently desirable to change the 
expression into its equivalent when y is made the independ- 
ent variable and x the function ; or to introduce some other 
variable of which both y and x are functions, and make it 
the independent variable. 

90. To Find the Values of ^, ^, ^L etc., 

dx dx 2 dx z 

when neither x nor y is Equicrescent. (Art. 55.) 

The value of the first derivative, -j-, will be the same 

whether x or y, or neither, is considered equicrescent. 

(Pit 
The value of the second derivative, -~ , was obtained in 

Art. 56 by differentiating -f- as a fraction with a constant 
denominator and dividing by dx. 

If we now consider that neither x nor y is equicrescent, 
and hence that both dx and dy are variables, and differen- 
tiate -j- , we have 



d ldy\ __ d 2 y dx — ffix dy ( . 

~dx\dx)~~^ d^ ~' U 



which is therefore the value of the second derivative when 
neither variable is equicrescent. 

Similarly, 

d?y _d L (d*y\ 
dx* ~ dx\dxV 

_ (dhj dx — d s x dy) dx — 3 (cPy dx — d?x dy) fix . . 
~ dtf ' ^ 

which is the value of the third derivative when neither 
variable is equicrescent, and so on for any other derivative. 



142 EXAMPLES. 

Cor. — If x is equicrescent, these equations are identical. 
If y is equicrescent, d 2 y = d s y = 0, and (1) becomes 
d 2 y d 2 x dy 



dx 2 dx 3 ' 

and (2) becomes 

d 3 y 3 (d 2 x) 2 dy — d z x dy dx 

dtf~~ ~~dx^ 



(3) 
(4) 



which are the values of the second and third derivatives 
when y is equicrescent. 

Sch. 1. — Hence, if we wish to change an expression when 
x is equicrescent into its equivalent where neither x nor y is 

d 2 t/ d 3 ti 
equicrescent, we must replace -~ % , ~- z , etc., by their com- 
plete values in (1), (2), etc.; but if we want an equivalent 
expression in which y is equicrescent, we must replace 

-r4, -j^ 3 , etc., by their values in (3), (4), etc. 

Sch. 2. — If we wish to change an expression in which x 

is equicrescent into its equivalent, and have the result in 

terms of a new independent variable t, of which # is a 

d 2 n d 3/ ii 
function, we must replace -— % , -~ , etc., by their complete 

values in (1), (2), etc., and then substitute in the resulting 
expression, in which neither x nor y is equicrescent, the 
values of x, dx, d% etc., in terms of the new equicrescent 
variable. 

examples. 

1. Transform x y£ + l-^J — ^ = 0, in which x is 

equicrescent, into its equivalents, (2) when neither x nor y 
is equicrescent, (2) when y is equicrescent. 



EXAMPLES. 143 

(1.) Replace y| by its value in (1), and multiply by dx 3 , 

CIX 

and we have 

x (d 2 y dx — d 2 x dy) + dif — dy dx 2 = 0. 

(£.) Put cPy = 0, divide by dy 3 to have the differential 
of the independent variable in its proper position, the de- 
nominator, and change signs, and we have 



d 2 x , (dx\z 



dif \dy 

d ii x dii ij 

2. Transform -^% — =— — - ~r + ., «, = 0, in which x 

dx 1 1 — x i dx 1 — x 2 

is equicrescent, into its equivalent when 6 is equicrescent, 

having given x = cos 0. 

■ dhi 
Eeplacing y^ by its complete value in (1), the given 

equation becomes 

d 2 y dx — d 2 x dy x dy y 

'dx 3 f^a* dx + T^x 2 "" 

x = cos 6; 

,\ dx = — sin 6 dd and d?x = — cos dd 2 . 
1 _ X 2 — s i n 2 e. 

Substituting, we have 

— d?y sin 6 dO + cos 6 dd 2 dy cos 6 dy y 

— sin 3 0tf^ + sin 2 6 sin dd + sin 2 ~" 

... ^ + y = o. (See Price's Calculus, Vol. I, p. 126.) 

n rr, a d 2 ll 1 dlj _ . . . , 

3. Transform yy 2 + - y-. + # = °? in which z is equi- 
crescent, into its equivalent, (1) when neither y nor 6 is 
equicrescent; (;?) when 6 is equicrescent; (£) when y is 
equicrescent, having given x 2 = 40. 



144 EXAMPLES. 

Eeplacing in this equation the complete value of -y^ , it 

T CtX 

becomes 

d 2 y dx — d 2 x dy 1 dy 

az 3 a; dx J 

x = 2(0)*; .\ dx = 6-^dd. 

26i 6* 

Substituting, we have 

(1.) ydd* + dydd 2 + d d 2 y dS — 6 dy d 2 6 = 0. 

(2.) When 6 is equicrescent, 6/ 2 (9 = ; therefore (1) be- 
comes 

y dd* + dydd 2 + Odhjdd = 0, 

(5.) When y is equicrescent, eft/ = ; therefore (1) be- 
comes 



cP8 /dd\* /cldf 



[i+ d £j 



4. Transform R = =■ into its equivalent, (1) in 

dx 2 
the most general form; (2) when 6 is equicrescent ; (3) 
.when r is equicrescent, having given x = r cos d, and 
?/ = r sin 0. 

The complete value of R is 

R= (dx* + clf)i 
d 2 xdy — d 2 ydx 

dx = dr cos — r sin <# 0, 
d?/ = dr sin + r cos d 0. 



EXAMPLES. 145 

d 2 x = cos 6 d 2 r — 2 sin 6 drdO — r cos 6 dd 2 

— r sin 6 cPd, 

d 2 y = sin 6 tpr -f 2 cos 6 drdO — r sin 6 dd 2 

+ r cos 6 d?6 ; 

.-. (<fc) 2 + (dy)* ~ dr 2 + rhW% 
d*xdy — d 2 ydx = reW0 — 2(W — rW — rtf>d 2 0. 

[rt> 2 + rW]* 



(1.) .-. * = 



(2.) J2 = - ^ 



rdVrfS ■+- 2tfrW9 + rW + rdnW 



(5.) 72 = 



^ 2 ^ «#' 



(See Serret's Calcul Differential et Integral, p. 94.) 
5. Transform 

(dif + dx 2 )* + adxePy = 0, 

in which x is equicrescent, into its equivalents, (1) when 
neither x nor y is equicrescent, (2) when y is equi crescent. 

(i.) (dy 2 + d* 2 )t + f / (d 2 ydx - d 2 xdy) = ; 

91. General Case of Transformation for Two Inde- 
pendent Variables. — Let « be a function of the inde- 
pendent variables, say u = f (x, y) : and suppose x and y 
functions of two new independent variables r and 6, 
so that, 

7 



And similarly 



146 EXAMPLES. 

3 = (r, 0) and y = xj, (r, 6) ; (1) 

then u may be regarded as a function of r and 0, through 

' du du 

x and y. It is required to find the values of -=- and -=- in 

cix ay 

terms of derivatives of u, taken with respect to the new 

variables r and 6. 

Since u is a function of r through x and y, we have 
(Art. 82), 

du _ du dx du dy 

dr ~~ dx dr dy dr ^ ' 

du du dx du dy m 

dd ~ dx"dd+ dy dd' W 

where the values of -y-, ~, -r-. ~^, can be found from (1). 
dr dr dd dd v ' 

Whenever equations (1) can be solved for r and d sepa- 
rately, we can find by direct differentiation the values of 

dr dr dd dd .. , . . ... ,. 

— , — , — , -=-, and hence by substituting m 

fix cly fix fly 

du du dr du dd 

dx dr dx dd dx ' 

n du du dr du dd 

and Ty = drdy + TeTy^ t ^> 

we can obtain the values of -r- and -= — 

dx dy 

When this process is not practicable, we can obtain their 
values by solving (2) and (3) directly, as follows : 

dx dx 

Multiply (2) by -^ and (3) by ^- and subtract ; then 

multiply (2) by ~ and (3) by -—- and subtract. We shall 
then have two equations, from which we obtain, 



du 


EXAMPLES. 

du dy du dy 
drdO~dddr 


dx 


dx dy 


dy dx' 




dr dd 
dudx 


dr dd 
die dx 


du 


dddr 


drdO 


dy 


dx dy 
drdd~ 


dy dx 
~drdd 



147 



(4) 



tin I /Jsr> rln § flni si nr * ' 



The values of — ' v y- 2 , etc., can be obtained from these, 

but the general formulae are too complicated to be of much 
practical use. (See Gregory's Examples, p. 35.) 

Cor. — If x = r cos and y = r sin 6, (4) and (5) 
become 

du __ du sin 6 du m du _ cos 6 du . du 

dx ~~ dr r dd ' dy r dd dr 



EXAMPLES. 



?L±_? to find du (Art 81). 

x — y x ' 

du == 2 ( xd y-y d x ) 

(x - yf 



2. u = sin ax + sin by + tan' 



il 



du = a cos axdx + b cos budy -\ \ — -^- - 

2/ 2 4- « 2 

3. m = sm l — 

y 

7 _ ydx — xdy 
y Vy 2 — x % 

4. u = sin (» + ?/). 

^ = cos (x + ?/) (tfc + dy). 



148 EXAMPLES. 



5. u- 



z 2 — a 2 



_ x (z 2 —a 2 ) (2ydx + xdy) — %xhjzdz 
(z 2 — a 2 ) 2 



6 u ~ lo • (% + ¥&- f \ 

\x — Vx 2 — v 2 / 



Vx 2 — y 2 

V <\/x 2 — y 2 

du 
dx 



du 
7. u = cot & to find ^- (Art. 82, Cor. 1). 



i-t-'i = — x y cosec 2 x y \- 4- log x ~)> 
\dx! U 6 dx! 

du 
8. w = sin («/ 2 — 2;), and y = log x, z = x 2 , to find -j- 

(Art. 82). ^ 

/<#?A __ 2 (?/ — a; 2 ) co s (y 2 — 2) 

x 



9. w = log tan 

5 y 



ldu\ 
Xdxl ~ 



y-*tx 



. X X 

sm - cos - 

y y 



10. u = log (x — a + V^ 2 — 2fl«). 

11. If w = £% 4 + <fy 2 2 3 4- x 2 y 2 z 2 , show that 

4i = 6 ^ 2 + 8 ^ (Art 83 -> 

12. If u — tan -1 — , show that 

V 1 -}- £ 2 + ?/ 2 

d 2 w 1 dhi 15xy 



dxdy (i + x % + ^f $e 2 ^ 2 (i + ^ + y 2)l 



EXAMPLES. 149 

13. u = x*y 2 + y 3 x 2 to find dhi (Art. 85). 
d 2 u = Gay^a* + ijxhjdxdy + 2# 3 ^a; 2 + Gxfdxdy 

+ Sxhjdxdy + 2a^7y 2 + Gxfdxdy + Qxhjdif 

= (6xif + 2# 3 ) <fo 2 + 12 (a% + ?/ 2 ^) My + (6a% + 2x z )dy 2 . 

14. xv + y_ ff = o. to find g. (Art. 87.) 



^ _ yaJ* -1 + y x log «/ 

15. a? — y* = 0. 



Ja; a^*" 1 -f- a* log a; 

fy _ f — x u l °g y 



dx x 2 — xy log x 

17. <* - Zaxy + f = 0. ^ = ^f^i. 

^ J ax y l — ax 

18. ye ax _ u. ^ ^ ^ + ^ 



19. a** + Vsec (ay) = 0. 

tfy _ y Vsec (ay) tan (xy) -+- 2a xy yx' J ~ 1 log # 

^ a; Vsec (ay) tan (xy) -f- 2«rtiP log a log a; 

20. a^ + 2ax*y - ay 3 - 0, to find ^ and g. 
(Ait. 88.) £ _ - __g-. 

g = _ [(iw _ 4^/) (2aa* - 3^ )2 

— 8«a; (4a? 4- 4aay) (2oa?>— 3cy 2 ) - (Ix* + 4=axyf Gay] 
-j. (9 fa - - 3a?/y = what ? 

Show that -j- = or ± y^2, when Z = and y = 0. 



150 EXAMPLES. 

21. Change the independent variable from x to t in 

d 2 y dy 
dx 2 dx 



(1 — x 2 ) -~ — x~- = 0, when x = cos £. 



,4 m. g = 0. 

22. Change the independent variable from x to 6 in 

t? + q — — » - 7 - + 7^ — ^-?to = 0, when x = tan fl. 
cfo 2 ^ 1 + z 2 ^ (1 + ^ 2 ) 2 

^«. J + y = o. 

23. Change the independent variable from x to r, and 
eliminate x, y, dx and J^, between 

. xdy — v^ a -. • z, 

z = — ^ -^-v , # = r cos (9. and y = r sin 0. 

An§. t = —: — 
dr 

24. Change the independent variable from x to z in 

P = x 2 y^ — 2# -- + 2/ j » wnen # = ^ an ^ y = ^( v + 2 )' 



. P = 6* 



g +( , + 1) j + , (y + 3) 




\^\> 



CHAPTER VIII. 

MAXIMA AND MINIMA OF FUNCTIONS OF A 
SINGLE VARIABLE. 

92. Definition of a Maximum and a Minimum. — If, 

while the independent variable increases continuously, a 
function dependent on it increases up to a certain value, 
and then decreases, the value at the end of the increase is 
called a maximum value of the function. 

If while the independent variable increases, the function 
decreases to a certain value and then increase*, the value at 
the end of the decrease is called a minimum value of the 
function. Hence,, a maxinbum value of a function of 
a single variable is a value which is greater than the 
immediately preceding and succeeding values, and a 
minimum value is less than the immediately pre- 
ceding and succeeding values. 

For example, sin increases as increases till the latter 
reaches 90°, after which sin <p decreases as increases ; 
that is, sin is a maximum when is 90°, since it is 
greater than the immediately preceding and succeeding 
values. Also, cosec decreases as increases till the latter 
reaches 90°, after which cosec increases as increases ; 
that is, cosec is a minimum when is 90°. since it is less 
than the immediately preceding and succeeding values. 

93. Condition for a Maximum or Minimum. — If y 

be any function of x, and y be increasing as x increases, 
the differential of the function is positive (Art. 12), and 

hence the first derivative -~ will be positive. If the func- 
tion be decreasing as x increases, the differential of the 



152 



GEOMETRIC ILLUSTRATION. 



function is negative, and hence the first derivative ~ will 
be negative. 

Therefore, since at a maximum value the function 
changes from increasing to decreasing, the first derivative 
must change its sign from plus to minus ; as the variable 
increases. And since, at a minimum value, the function 
changes from decreasing to increasing, the first derivative 
must change its sign from minus to plus. But as a function 
which is continuous* can change its sign only by passing 
through or oo , it follows that the only values of the 
variable corresponding to a maximum or a minimum 
value of the function, are those ivhich make the first 
derivative O or oo . 

94. Geometric Illustra- 
tion. — This result is also 
evident from geometric con- 
siderations ; for, let y = f(x) 
be the equation of the curve 
AB. At the points P, P', P", 
P'", the tangents to the curve 
are parallel to the axis of x, 
and therefore at each of these points the first derivative 
/' (x) = 0, by Art. 56 a . 

We see that as x is increasing and y approaching a 
maximum value, as PM, the tangent to the curve makes 
an acute angle with the axis of x ; hence, approaching P 

At P the tangent becomes parallel 




M M' M" M"iVT 



Fig. 9. 



from the left - 7 - is 
ax 



to the axis of x ; hence, 



dy 

dx 



0. Immediately after pass- 



ing P the tangent makes an obtuse angle with the axis 

of x ; hence, -- is — . 
ax 



* In this discussion the function is to be regarded as continuous. 



CRITERION OF MAXIMA AND MINIMA. 153 

Also in approaching a minimum value, as P'M', from the 
left, we see that the tangent makes an obtuse angle with 

the axis of z. and hence - 7 ~ is — . At the point P', -- = 0. 
dz J ' dz 

After passing P, the angle is acute and -.- is -(- . 

In passing P"', -j- changes sign by passing through go , 

P"M'" is a minimum ordinate. In approaching it from the 

left the tangent makes an obtuse angle with the axis of x, 

dy 
and hence -j- is — . At P" the tangent is perpendicular 

az 

dy 
to the axis of z, and -^- = co . After passing P'"M" ; , the 

angle is acute and ~ is + . 

While the first derivative can change its sign from + to 

— or from — to -f only by passing through or go , it 

does not follow that because it is or oo, it therefore 

necessarily changes its sign. The first derivative as the 

variable increases may be + , then 0, and then + , or it may 

be — , then 0, and then — . This is evident from Fig. 9, 

where, at the point D, the tangent is parallel to the axis of 

dy 
z, and -- is 0, although just before and just after it is — . 

Hence the values of the variable which make -.- = or co , 

dz 

are simply critical* values, i. e., values to be examined. 

As a maximum value is merely a value greater than that 

which immediately precedes and follows it, a function may 

have severed mazimum values, and for a like reason it may 

have several minimum values. Also, a maximum value 

may be equal to or even less than a minimum value of the 

same function. For example, in Fig. 9, the minimum P'M' 

is greater than the maximum P iv M iv . 

* See Price's Cal., Vol. I, p. 237, 



154 CONDITION GIVEN BY TAYLOR'S THEOREM. 

95. Method of Discriminating between Maxima 
and Minima. — Since the first derivative at a maximum, 
state is 0, and at the immediately succeeding state is — , it 
follows that the second derivative, which is the difference 
between two consecutive first derivatives,* is — at a maxi- 
mum. Also, since the first derivative at a minimum state 
is 0, and at the immediately succeeding state is + , it fol- 
lows that the second derivative is -f at a minimum. There- 
fore, for critical values of the variable, a function is at a 
maximum or a minimum state according as its 
second derivative at that state is — or -f . 

96. Condition for a Maximum or Minimum given 
by Taylor's Theorem. — Let u =f(x) be any continuous 
function of one variable ; and let a be a value of x corre- 
sponding to a maximum or a minimum value of f(x). 
Then if a takes a small increment and a small decrement 
each equal to h, in the case of a maximum we must have, 
for small values of h, 

f(a) >f(a + h) and f(a) > f(a - ft) ; 

and for a minimum, 

f(a)<f(a + h) and f(a) <f(a- h). 

Therefore, in either case, 

f(a + k)--f.(a) and f(a-h)-f(a) 
have both the same sign. 

By Taylor's Theorem, Art. 66, Eq. 8, and transposing, 
we have 

/ (a + h) -/(«)= /' (a) h + /" (a) |" + etc. ; (1) 

7,2 

f(a - h) -/(«) = -/' (a) h +/" (a) f - etc. (2) 



* Remembering that the first value is always to be subtracted from the second. 



FINDING MAXIMA AND MINIMA VALUES. 155 

Now if h be taken infinitely small, the first term in the 
second member of each of the equations (1) and (2) will be 
greater than the sum of all the rest, and the sign of the 
second member of each will be the same as that of its first 
term, and hence f(a + h) —f(a) and f(a — h) —f(a) 
cannot have the same sign unless the first term of (1) and 
(2) disappears, which, since h is not 0, requires that 
/'(«) = 0. 

Hence, the values of x which make f(x) a maxi- 
mum or a minimum are in general roots of the equa- 
tion, /' (x) = 0. 

Also, when /' (a) = 0, the second members of (1) and 
(2), for small values of li> have the same sign as / (a) ; 
that is, the first members of (1) and (2) are both positive 
when /" (a) is positive, and negative when f" (a) is nega- 
tive. Therefore, f(a) is a maximum or a minimum 
according as f" (a) is negative or positive. 

If, however, /" (a) vanish along with /' («), the signs of 
the second members of (1) and (2) will be the same as 
/'" (a), and since /'" (a) has opposite signs, it follows that 
in this case f(ci) is neither a maximum nor a mini- 
mum unless f" (a) also vanish. But if /'" (a) = 0, 
then f(a) is a maximum when f w (a) is negative, and a 
minimum when f' ,v (a) is positive, and so on. If the first 
derivative which does not vanish is of an odd order, f(a) is 
neither a maximum nor a minimum ; if of an even order, 
f(a) is a maximum or a minimum, according as the sign 
of the derivative which does not vanish is negative or posi- 
tive. 

97. Method of Finding Maxima and Minima 
Values. — Hence, as the result of the preceding investiga- 
tion we have the following rule for finding the maximum 
or minimum values of a given function, f(x). 

Find its first derivative, f (x) put it equal to 0, 
and solve the equation thus formed, f (x) == 0. Sub- 



156 MAXIMA AND MINIMA VALUES ALTERNATE. 

stitute the values of x thus found for x in the second 
derivative, f" (as). Each value of x which makes 
the second derivative negative will, when substituted 
in the function f(x) make it a maximum; and each 
value which makes the second derivative positive will 
make the function a minimum. If either value of 
x reduces the second derivative to 0, substitute in the 
third, fourth, etc., until a derivative is found which 
does not reduce to 0. If this be of an odd order, the 
value of x will not make the function a maximum 
or minimum ; but if it be of an even order and nega- 
tive, the function will be a maximum ; if positive, a 
minhnum. 

Second Rule. — It is sometimes more convenient to 
ascertain whether a root a of f" (x) = corresponds to a 
maximum or a minimum value of the function by substi- 
tuting for x; in /' (x), a — li and a -j- h, where h is infini- 
tesimal. If the first result is -f- and the second is — , 
a corresponds to a maximum ; if the first result is 
— and the second is -f, it corresponds to a minimum. 
If both results have the same sign, it corresponds to 
neither a maximum nor a minimum. (See Arts. 93, 94.) 

98. Maxima and Minima Values occur alternately. 

— Suppose that f(x) is a maximum when x = a, and also 
when x = b, where b > a; then, in passing from a to b, 
when x = a + h (where h is very small), the function is 
decreasing, and when x — b — li, it is increasing; but in 
passing from a decreasing to an increasing state, it must 
pass through a minimum value ; hence, between two maxi- 
ma one minimum at least must exist. 

In the same way, it may be shown that between two 
minima one maximum must exist. 

This is also evident from geometric considerations, for in 
Fig. 9 we see that there is a maximum value at P, a mini- 
mum at P', a maximum at P", a minimum at P" ; , and so on. 



APPLICATIONS OF AXIOMATIC PRINCIPLES. 157 

99. The Investigation of Maxima and Minima is 
often facilitated by the following Axiomatic Prin- 
ciples : 

1. If u be a maximum or minimum for any value of x, 
and a be a positive constant, an is also a maximum or mini- 
mum for the same value of x. Hence, before applying the 
rule, a constant factor or divisor may be omitted. 

2. If any value of x makes u a maximum or minimum, 
it will make any positive power of u a maximum or mini- 
mum, unless u be negative, in which case an even power of 
a minimum is a maximum, and an even power of a maxi- 
mum is a minimum. Hence, the function may be raised, 
to any power ; or, if under a radical, the radical may 
be omitted. x 

3. Whenever u is a maximum or a minimum, logic is a 
maximum or minimum for the same value of x. Hence, 
to examine the logarithm of a function ire have only 
to examine the function itself. When the function con- 
sists of products or quotients of roots and powers, its exam- 
ination is often facilitated by passing to logarithms, as the 
differentiation is made easier. 

4. When a function is a maximum or a minimum, its 
reciprocal is at the same time a minimum or a maximum; 
this principle is of frequent use in maxima and minima. 

5. If u is a maximum or minimum, u ± c is a maximum 
or minimum. Hence, a constant connected by -f- or — 
may be omitted. 

Other transformations are sometimes useful, but as they 
depend upon particular forms which but rarely occur, they 
may be left to the ingenuity of the student who wishes to 
simplify the solution of the proposed problem. 

It is not admissible to assume x = oo in searching for 
maxima and minima, for in that case x cannot have a suc- 
ceeding value. 



158 EXAMPLES. 



EXAMPLES. 

1. Find the values of x which will make the function 
u = Qx + 3x 2 — 4a: 3 a maximum or minimum, and the cor- 
responding values of the function u. 

Here ^ =_- 6 + 6z - 12a* 

Now whatever values of x make w a maximum or mini- 

die 
mum, will make -j- = (Art. 97) ; therefore, 

6 + 6x — 12z 2 = 0, or x* — \x = -| ; 
.-. « = J ± f = + 1 or - f 

Hence, if u have maximum or minimum values, they 
must occur when x = 1 or — \. 

To ascertain whether these values are maxima or minima, 
we form the second derivative of u ; thus, 

£ = .-•* 

dHi 
When x = 1, y^ = — 18, which corresponds to a maxi- 
mum value of u. 

clht 
When x = — J-, ---== + 18, which corresponds to a 

minimum value of w. 

Substituting these values of x in the given function, we 
have 

When x = 1, u =6 + 3 — 4 = 5, a maximum. 

When a; == — -J, u ■ = — 3 + } + J = — J, a minimum. 

2. Find the maxima and minima values of u in 

tt — a* — 8a? + 22a? — 24a: + 12. 

*? = 4^3 _ 24a;2 + 44a; — 24 = 0. 
dx 



EXAMPLES. 159 

or, a? — 6a? + \lx — 6 = 0. 

By trial, a: = 1 is found to be a root of this equation : 
therefore, by dividing the first member of this equation by 
x — 1, we find for the depressed equation, 

a* _ ox + 6 = 0; ,\ a? == 2 or 3. 

Hence the critical values are x = 1, x = 2, and # == 3. 

V± = 12a? — 48a; + 44 = + 8, when x = 1. 

= — 4, when x = 2. 

= -f 8, when £ = 3. 
Therefore we have, 

when x = 1, w = 3, a minimum ; 

when x = 2, u = 4, a maximum ; 

when a; = 3, it = 3, a minimum. 

3. Find the maxima and minima values of u in 
u = (x - l) 4 (x + 2) 3 . 

^ = 4 (x - l) 3 (» 4- 2) 3 + 3 (a; 4- 2f(x - l) 4 
= (a! _ i)3 (., + 2)2 [4 (x 4- 2) + 3 (* - 1)], 
or g = (s - l) 3 (x + 2)2 (7* + 5) = ; (1) 

.-. (a; — 1) == 0, (x 4- 2) = 0, (7a? 4- 5) = 0. 

/. a? == 1, a; = — 2, and x = ■— f, as the critical values 
of a;. 

In this case, it will be easier to test the critical values by 

du 
the second rule of Art, 97; that is, to see whether -3- 

dx 

changes sign or not in passing through x = 1, — 2, and 

— -| in succession. 



160 EXAMPLES. 

If we substitute in the second member of (1), (1 — h) 
and (1 -f- h) for x, where h is infinitesimal, we get 

g =(l-h- 1)3(1 -h + 2)«'[7 (1 - ft) + 5] 
= _ &a ( 3 _ fry (!2 _ 7ft) = _. 

and ^ = (1 + 7* - 1) 3 (1 + h + 2)* [7(1 + A) + 5] 

= A 3 (3 + 7;) 2 (12 + 7/0 = + 

Therefore, as -=- changes sign from — to + at x = 1, 
the function u at this point is a minimum. 

When x = — 2, — does not change sign ; .-. u has no 
maximum or minimum at this point. 

When x = — -f-, — changes sign from -f to — ; .-. u, 
at this point, is a maximum. 

Hence, when x == 1, w = 0, a minimum. 

124.93 

It is usually easy to see from inspection whether — 

cix 

changes sign in passing through a critical value of x, with- 
out actually making the substitution. 

4. Examine u = b + (x — a) 3 for maxima and minima. 

clu 

— = 3 (x — a) 2 = ; .'. x = a, and u = b. 

dhc 
Since a; = a makes -^ = 0, we must examine it by the 

clu 
second rule of Art. 97, and see whether -=- changes sign at 

x = a. 



when ^ = — f , u = — — — , a maximum. 



EXAMPLES. 161 

— = 3 (a — h — a) 2 = 37i 2 is the value of -5- immediately 
preceding x = a. 

— 1 = 3 (« + h — df = 3h 2 is the value of -=- immediately 



Therefore, as -r- does not change sign at x = a, u = h 
dx ° c 

is neither a maximum nor a minimum. 

5. Examine u = b + (x — a) 4 for maxima and minima. 

-=- = 4 (2; — ff) 3 = ; •'• * = rt and ?< = b. 
dx v ' ' 

It is easy to see that y- changes sign from — to + at 

x = a ; .\ x = a gives a = d, a minimum. 

(2; _j_ 2) 3 

6. Examine w = 7- ^4 for maxima and minima. 



or 00; 



rfw _ (a; + 2 )2 (s — 13 ) 
^ ~" (a; - 3) 3 

.-. a? — — 2, 13, or 3. 

"We see that when x == — 2, -— does not change sign ; 
.'. no maximum or minimum ; 

when x = 13, -=- changes sign from — to + ; 

.*. a minimum ; 

when x = 3, -=- changes sign from + to — , 

.*. a maximum; 
hence when x = 13, w = if-% a minimum value ; 
and when 2 = 3, u = cc , a maximum value. 



162 EXAMPLES. 

7. Examine u = b + (x — a)* for maxima and minima. 

— - = 4 (x — a)* = ; .*. x = a and u = b. 
dx 3K ' 

When x = a, -r- changes sign from — to +. 

x = a gives u = b, a minimum. 

8. Examine u = b — (a — x)% for maxima and minima. 

— - = f (a — x)% = ; /. .x = a and u = b. 

cJu 
When x = a, -j- changes sign from -f to — . 

.♦. x — a gives u = b, a maximum. 

9. Examine u — b + tya s — 2a 2 x + ax 2 for maxima and 
minima. 

If u is a maxima or minima, u — b will be so ; therefore 
we omit the constant b and the radical by Art. 99, and get 

u' = a? — 2a 2 x + ax 2 ; 

flit' 

-— =: —2a 2 -\- %ax = 0; .*. x = a and u = b. 

ax 

(In 
When x = a, -=-= changes sign from — to -f. 
ax 

.-. x = a gives « =: #, a minimum. 

a 2 x 

10. Examine w = -. — .-= for maxima and minima. 

(a — x) 2 

Using the reciprocal, since it is more simple, and omitting 
the constant a 2 (Art. 99), we have 

, (a — x) 2 a 2 

u = = 2a 4- x ; 

a; # 

dw' a 2 , i , dV 2a 2 

- ^ = -^ +1 = ' and S^** 



EXA31PLES. 163 

dhc' , 2 
... x=±a, and /. -^ = ± -. 

Hence, x = + « makes w' a minimum, and a; = — a 
makes it a maximum; therefore, since maxima and minima 
values of u' correspond respectively to the minima and 
maxima values of u (Art. 99, 4), we have, 

when x = a, u = qo , a maximum. 
" x = — a, u = — - , a minimum. 

Find the values of oc which give maximum and 
minimum values of the following functions : 

1. u = x s — 3x 2 — 24a? + 85. 

Ans. x = — 2, max. ; x = 4, min. 

2. u = 2x* — 21a; 2 -f 36a? — 20. 

a? = 1, max. ; a; = 6, min. 

3. w = x B — 18a; 2 + 96a; — 20. 

x = 4, max. ; x = 8, min. 

, (a — x) 2 

4. w = --• x = ±a, mm. 

a — 2x 4 

K 1 + SX 

5. w = — - • # = — 1 T 4 X , max. 

6. ^ — ^s _ 3^ _ 9a . + 5# 

a == — 1, max. ; x = 3, min. 

7. w = a* _ 3^2 + 6a? + 7. 

Neither a max. nor a min. 

8. u = (a; — 9) 5 (a; — 8) 4 . 

a; = 8, max. ; x = 8f , min. 

9. u = x = cos a-, max. 

1 + x tan x 

10. u = sin 3 a; cos a?, x = 60°, max. 

11. u = — — r a; ±= 45 , max. 

1 + tan x 



164 



GEOMETRIC PROBLEMS. 



12. u = sin x -+- cos x. . 

x = 45°, max. ; x = 225°, min. 

log x 



13. w = 



x" 



e n , min. 



GEOMETRIC PROBLEMS. 

The only difficulty in the solution of problems in maxima and 
minima consists in obtaining a convenient algebraic expression for the 
function -whose maximum or minimum value is required. No gen- 
eral rule can well be given by which this expression can be found. 
Much will depend upon the ingenuity of the student. A careful ex- 
amination of all the conditions of the problem, and tact in applying 
his knowledge of principles previously learned in Algebra, Geometry, 
and Trigonometry, with experience, will serve to guide him in form- 
ing the expression for the function. After reducing the expression to 
its simplest form by the axioms of Art. 99, he must proceed as in 
Art. 97. 



1. Find the maximum cylinder which 
can be inscribed in a given right cone 
with a circular base. 

Suppose a cylinder inscribed as in 
the figure. Let AO == b, DO = a, 
CO = x, CE = y. 

Then, denoting the volume of the 
cylinder by v, we have 

V = ny 2 x. 




Fig. 10. 



(1) 



From the similar triangles DOA and DCE, we have 
DO : AO :: DC : EC, 
or a : b : : a — x : y : 



which in (1) gives 



v = 7t— _ (a — xfx. 
a 2 v 



(2) 



GEOMETRIC PROBLEMS. 165 

Dropping constant factors (Art. 99), we have 
= (a — x) 2 x = a 2 x — 2ax 2 -f x z ; 
— a 2 _ ± ax + 3^2 _ o ? 

.-. x = a or \a. 



u 
du 
dx 



or x 2 — %ax = — Jfl 2 ; 
dhb . a 

= 2a, when x = a, .'. minimum ; 
= — 2a, when x = |#, .% maximum. 

Hence the altitude of the maximum cylinder is one-third 
of the cone. 

The second value of x in (2) gives 

Volume of cone = ^~al 2 . 
.-. Volume of cylinder = f volume of cone. 



y = - (« - 4«) = f* 



•adius of base of cylinder. 



2. What is the altitude of the 
maximum rectangle that can be in- 
scribed in a given parabola ? 

Let AX = a, AH = x, DH = y 



/ 



M X 

Fig. [I, 



and ^1 = area of rectangle. Then L . 

we have 

A = 2y (a — x). 

But from the equation of the parabola, we have 

y = \/2px, 
which in (1) gives A = 2\/2p:r (a — x). 




c 

a) 



(2) 



u' = Vx(a 

du' 
dx 



= \ax~^ 



x) = ax 1 * — x 2 . 

|#2 = o. .-. x = %a. 




166 GEOMETRIC PR0BLE3TS. 

Since this value of x makes y- change sign from + to 

— , it makes the function A a maximum; therefore the 
altitude of the maximum rectangle is § a. 

3. What is the maximum cone that 
can be inscribed in a given sphere ? 

Let ACB be the semicircle, and 
A CD the triangle which, revolved Fl l2t 

about AB, generate the sphere and 
cone respectively. Let AO = r, AD = x, and CD = y, 
and v = volume of cone. 

Then v = \mfx. (1) 

But if = AD xDB = (2r — x) x, 

which in (1) gives v = In (2r — x) x% (2) 

or u = 2rx 2 — x z ; 

rj?/. 

,\ ~ = \rx — 3x* = 0. 
ax 

.*. x = and fr. 

The latter makes du change sign from -f to — ; .*. it 
makes v a maximum. 

Hence the altitude of the maximum cone is f of the 
diameter of the sphere. 

The second value of x in (2) gives 

v = ±n (2r - 4f) (irY = ffwr" = A X |fl»* 

Volume of sphere = f^r 3 ; 

•\ the cone = fc of the sphere. 

4. Find the maximum parabola which can be cut from a 
given right cone with a circular base, knowing that the area 
of a parabola is § the product of its base and altitude. 



GEOMETRIC PROBLEMS. 



167 



Let AB =:«,AC = b, and BH = x; 
then AH = a — x. 



FE = 2EH = 2VAHxBH 



= 2V(« — x)x. 
Also, BA:AC :: BH : HD, 



or 



« : b : : a; : HD = - x. 

a 



Calling the parabola A, Ave have 




A = f FE x HD l= = i-xV{a — x) x, 



or 



u = ax 6 — x\ 



du 

dx 



Sax 2 — 4a 3 = ; 
x = and a; = f-fl. 
The second value makes -j- change sign from -f to — , 

(.IX 

and .-. makes the function A a maximum. 



.-. A = i---iaV(a - la) la = \ab\/Z, 

which is the area of the maximum parabola. 

Eem. — In problems of maxima and minima, it is often more con- 
venient to express the function u in terms of two variables, x and y, 
which are connected by some equation, so that either may be regarded 
as a function o^ the other. In this case, either variable of course may 
be eliminated, and u expressed in terms of the other, and treated by 
the usual process, as in Examples 1, 2, and 3 It is often simpler, 
however, to differentiate the function u, and the equation between x 
and y, with respect to either of the variables, x, regarding the other, 

y, as a function of it. and then eliminate the first derivative. -¥-. The 

dx 

seccnd method of the following example will illustrate the process. 



168 



GEOMETRIC PROBLEMS. 



5. To find the maximum rectangle inscribed in a given 
ellipse. 

Let CM == x, PM = y, and 
A = area of rectangle. Then we 
have 

A — Axy, (1) 

and a 2 if + b 2 x 2 = a 2 b 2 . (2) 



F 


s 




P 


./ 


m' 


C M 


\ 


\ 






/ 




v_ 


^/ 





1st Method. — From (2) we get 



y = - Vfl 2 — x 2 , 
J a 

b 



Fig. 14. 



which in (1) gives A = 4 - x Va 2 — x 2 , 
or u = a 2 x 2 — x*. 

rloi 

— 2a 2 x — 4t 3 = 0. 



dx 



v% 



rr = H = makes -=- change sign from -f to — ; .'.it 

makes A a maximum. 

Hence, the sides of the maximum rectangle are a V% 
and b V%, and the area is 2ab. 

2d, Method. — Differentiate (1) and (2) with respect to x 
after dropping the factor 4 from (1), and get 



dA , dy 

dx- = y + x Tx = °> 


•' 


^ — _ y. 

dx x 


2a 2 y C ^ + 2¥x = 0; 




dy b 2 x 
dx ~~ a 2 y 




or 


b 2 x 2 = ahf ; 



which in (2) gives 



GEOMETRIC PROBLEMS. 169 

2a 2 y 2 — a 2 b 2 , :. y = — = and x = — =• 
V2 V'2 

6. Find the cylinder of greatest convex surface that 

can he inscribed in a right circular cone, whose altitude 

is h and the radius of whose base is r. „ nkr 

ourtace = — -• 

7. Determine the altitude of the maximum cylinder 
which can he inscribed in a sphere whose radius is r. 

Altitude = §/• VS. 

8. Find the maximum isosceles triangle that can be 
inscribed in a circle. An equilateral triangle. 

9. Find the area of the greatest rectangle that can be 
inscribed in a circle whose radius is r. 

The sides are each r a/2. 

10. Find the axis of the cone of maximum convex sur- 
face that can be inscribed in a sphere of radius r. 

The axis = Jr. 

11. Find the altitude of the maximum cone that can be 
inscribed in a paraboloid of revolution, whose axis is a, the 
vertex of the cone being at the middle point of the base of 
the paraboloid. Altitude = ^a. 

VI. Find the altitude of the cylinder of greatest convex 
surface that can be inscribed in a sphere of radius r. 

Altitude = >• a 2. 

13. From a given surface s, a vessel with circular base 
and open top is to be made, so as to contain the greatest 
amount. Find its dimensions. (See Remark under Ex. 4.) 

e 

The altitude = radius of base = \ - — 

On 

14. Find the maximum cone whose convex surface is 
constant. The altitude = V 2 times the radius of base. 

15. Find the maximum cylinder that can be inscribed in 
an oblate spheroid whose semi-axes are a and b. 

The radius of base = a V-l ; the altitude = b 

V3 



170 EXAMPLES. 

16. Find the maximum difference between the sine and 
cosine of any angle. When the angle = 135°. 

17. Find the number of equal parts into which a must 
be divided so that their continued product may be a 
maximum. 

Let x be the number of parts, and thus each part equals 

-, and therefore u = l-\ , from which we get x = - ; 

therefore each part = e, and the product of all = (e)' e . 

18. Find a number x such that the xth root shall be a 
maximum. x = e = 2.71828 -f-. 

19. Find the fraction that exceeds its m th power by the 
greatest possible quantity. / 1 \-I T 

\m I 

20. A person being in a boat 3 miles from the nearest 
point of the beach, wishes to reach in the shortest time a 
place 5 miles from that point along the shore ; supposing 
he can walk 5 miles an hour, but row only at the rate of 
4 miles an hour, required the place he must land. 

One mile from the place to be reached. 

21. A privateer wishes to get to sea unmolested, but has 
to pass between two lights, A and B, on opposite head- 
lands, the distance between which is c. The intensity of 
the light A at a unit's distance is a, and the intensity of B 
at the same distance is b ; at what point between the lights 
must the privateer pass so as to be as little in the light as 
possible, assuming the principle of optics that the intensity 
of a light at any distance equals its intensity at the distance 
one divided by the square of the distance from the light. 

a* + $ 

22. The flame of a candle is directly over the centre of a 
circle whose radius is r ; what ought to be its height above 
the plane of the circle so as to illuminate the circumfer- 
ence as much as possible, supposing the intensity of the 



EVALUATION. 



171 




Fig. 15. 



light to vary directly as the sine of the angle under which 
it strikes the illuminated surface, and inversely as the 
square of its distance from the same surface. 

Height above the plane of the circle == r \/\ 

23. Find in the line joining the centres of two spheres, 
the point from which the greatest 

portion of spherical surface is 
visible. 

The function to be a maximum 
is the sum of the two zones whose 
altitudes are AD and ad; hence 
we must find an expression for the areas of these zones. 

Put CM = R and cm = r, Cc = a and CP = x. 

The area of the zone on the sphere which has R for its 
radius (from Geometry, or Art. 194) = 2-RAD = 2~R 2 

/ R 3 \ 
— 2tRCD = 27T (R 2 — — j, and in the same way for the 

other zone, from which we readily obtain the solution. 

_ am 

24. Find the altitude of the cylinder inscribed in a sphere 
of radius r, so that its whole surface shall be a maximum. 

i 



Altitude = r [~2 (l - -V) 



CHAPTER IX. 



TANGENTS, NORMALS AND ASYMPTOTES. 



100. Equations of the Tangent and Normal. — Let 

P, (x\ y') be the point of tangency ; 
the equation of the tangent line at 
(x', y') will be of the form (Anal. 
Geom., Art. 25) 

y — y = a(x —x'), (1) 

in which a is the tangent of the 

angle which the tangent line makes 

with the axis of x. It was shown in 

Article 56a that the value of this tangent is equal to the 

derivative of the ordinate of the point of tangency, with 

respect to x, 

dx' 




or 



Hence 



y-y 



dy> 



dx 



7 (^ — %'), 



(2) 



is the equation of the tangent to the curve at the point 
{x\ y'), x and y being the current co-ordinates of the 
tangent. 

Since the normal is perpendicular to the tangent at the 
point of tangency, its equation is, from (2), 



, dx' , ,. 



(3) 



(Anal. Geom., Art. 27, Cor. 3.) 



EXAMPLES. 173 

Rem. — To apply (2) or (3) to any particular curve, we 

dii' dx' 
substitute for -^-, or -r~, , its value obtained from the equa- 
tion of the curve and expressed in terms of the co-ordinates 
of the point of tangency. 

EXAMPLES. 

1. Find the equations of the tangent and normal to the 

ellipse 

ahf + Vx> = aW. 

W find ^ — - — • • — - - — • 

dx ~ a z y' dx' ~ o?y" 

and this value in (2) gives, 

&x' . , x 

y ~ V = ~^' {X ~ X); 

which by reduction becomes, 

ahjij + b 2 xx' = aW, 
which is the equation of the tangent ; and 

ahf ■ n 

y-y =w {x ~ x) 

is the equation of the normal. 

2. Find the equations of the tangent and normal to the 
parabola y 2 = 2px. 

We find * - P ■ & - * 







dx " 


~y' 


" dx' ~ y" 


and this 


value 


in (2) 


gives 








y 


— y' = 


= }(*-*), 


or 




yy' - 


-y'*-- 


= px — px'. 


But 






y' 2 - 


- 2px' ; 



174 EXAMPLES. 

••• yy' = p (z + x'\ 

which is the equation of the tangent ; and 

/ y' / /\ 

2/ - y = - - (a - * ) 
is jf/w equation of the normal. 

3. Find the equations of the tangent and normal to an 
hyperbola. 

Tangent, a 2 yy' — b 2 xx' = — a 2 b 2 . 

Normal, y — y' = — ^ (# — a;'), 

4. Find the equation of the tangent to 3# 2 + x 2 — 5 = 0, 
at x = 1. 

Here -^ = — §-, = ro^H = H 1 - 29 about, 

<&k 3?/ ± 3.465 

which in (2) gives 

yT 1.155 == HF .29(3 — 1), 

or y = =F .29z ± 1.44. 

Hence there are two tangents to this locus at 3 = 1, 
their equations being 

y = — .293 + 1.44 and y = + .29a; — 1.44. 

5. Find the equation of the tangent to the parabola 
y 1 = 9x, at x = 4. 

At (4, 6) the equation is y = £3 + 3. 
« (4, — 6) " " " y=— i® — 3. 

6. Find the equation of the normal to y 2 = 2x 2 — x% at 

3 = 1; 

At (1, +1) the equation is # = 2x — 3. 
« (i ? _ i) « « « y — — 2x + 3. 

7. Find the equation of the normal to y 2 = 63 — 5, at 
y = 5, and the angle which this 'normal makes with the 
axis of x. y — — i% + >t" 5 an g le = tan " _1 ( _ I)* 



LENGTH OF TANGENT, NORMAL, ETC. 



175 



101. Length of Tangent, Normal, Subtangent, 
Subnormal, and Perpendicular on the Tangent from 
the Origin. 

Let PT represent the tangent at 
the point P, PN the normal; draw 
the ordinate PM ; then 

MT is called the subtangent, 

MN " " " 'subnormal. 

Let « = anglePTM; 

then tan « = -£-, (Art. 56a). 

1st. TM = MP cot a = y'~\ 

dx 

.*. Subtangent == y'.-j-r 

MN = MP tan MPN = y tan «; 




2d. 



3d. 



Subnormal = y 



clx' 



PT = VPM? + MT 2 



=v 



y? + 



«V\ 8 



/ , dx \ 



4th. 



PN = VPM S + MX* 



Normal = y'\ / 1 + f-^j • 
5th. The equation of the tangent at P (a?', y') is (Art. 100), 



176 EXAMPLES. 

dy' , , s 

y- y = *?<*-*>» 

or xdy' — ydx — x'dy' + ?/W = ; 

which, written in the normal form, is 

xdy'-yd_x^ z xd l+j 1 Uy ^ 

hence, OD ±z y ' dx _ ~ ^ L ( An al. Geom., Art. 24). 

V(tfs') 2 + W) 2 

,: Perpendicular on the tangent from the origin 

y'dx' — x'dy' 

Sch. — In these expressions for the subtaugent and sub- 
normal it is to be observed that the subtangent is measured 
from M towards the left, and the subnormal from M towards 

the right. If, in any curve, y' yA is a negative quantity, it 

denotes that "N lies to the left of M, and as in that case 

y\-j~, is also negative, T lies to the right of M. 

EXAMPLES. 

1. Find the values of the subtangent, subnormal, and 
perpendicular from the origin on the tangent, in the ellipse 

Here C K - _ h l x l. 

±lere dx' ~ ahf 

dx d^ti 2 

Hence, the subtangent = y' -^-, = — -^-r , 

dii' ffi 

the subnormal = y' ' -p, — %' ; 

ax a 



EXAMPLES. 



177 



the perpendicular from origin on tangent 

- aW 

~ (ahj'* + Vx'»fi 

2. Find the snbtangent and subnormal to the Cissoid 

a? 



sr 



2a — x 



(See Anal. Geom., Art. 149.) 



dx' 



Here 

Hence, the sub tan gent 



the subnormal = 



x? (3a — x) 

~(2a^~x~) J 

x (2a — x) 
3a — x 

x* (3a — x ) 
(2a - xf ' 



3. Find the value' of the subtangent of y 2 = 3a; 2 — 12, 

at x = 4. Subtangent = 3. 

4. Find the length of the tangent to y 1 = 2x, at a; = 8. 

Tangent = 4VT?. 

5. Find the values of the normal and subnormal to the 
cycloid (Anal. Geom., Art. 156). 

__ H 

x = r vers -1 - — \/2ry — y 2 ; 



dx _ y _ V2ry—f -4J^ 

dy ~ VZry^y 2 ~ 2r — y 

dy _ 2r-y 

dx 




Fig. 18. 



V2ry — if 

,\ Subnormal = V2ry — f = MO. 

Normal = \/2ry = PO. 

It can be easily seen that PO is normal to the cycloid at 
P ; for the motion of each point on the generating circle at 



178 



POLAR CURVES. 



the instant is one of rotation about the point of contact 0, 
i. e., each point for an instant is describing an infinitely 
small circular arc whose centre is at ; and hence PO is 
normal to the curve, i.e., the normal passes through the 
foot of the vertical diameter of the generating circle. Also, 
since OPH is a right angle, the tangent at P passes through 
the upper extremity of the vertical diameter. 

6. Find the length of the normal in the cycloid, the 
radius of whose generatrix is 2, at y = 1 . Normal = 2. 



POLAR CURVES, 

102. Tangents, Normals, Subtangents, Subnor- 
mals, and Perpendicular on Tangents. 

Let P be any point of the 
curve APQ, the pole, OX the 
initial line. Denote XOP by 
0, and the radius-vector, OP, 
by r. Give XOP the infinitesi- 
mal increment POQ = dd, then 
OQ = r + dr. From the pole 
0, with the radius OP = r, de- 
scribe the small arc PR, sub- 
tending dd ; then, since dd = ad 
is the arc at the unit's distance 
from the pole 0, we have 




Fig. 19 



PE = rdd and EQ = dr. (1) 

Let PQ, the element * of the arc of the curve, be repre- 
sented by ds. 

.-. PQ 8 = PR 2 + EQ 2 , 

or ds 2 = r*dd 2 + dr 2 . (2) 

Pass through the two points P and Q the right line QPT; 



* See Art. 56a, foot-note. 



POLAR C UR VES. 1 \ 9 

then, as P and Q are consecutive points, the line QPT is a 
tangent to the curve at P (Art. 56«). Through: P draw the 
normal PC, and through draw COT perpendicular to OP, 
and OD perpendicular to PT. The lengths PT and PC are 
respectively called the polar tangent and the polar normal. 
OC is called the polar subnormal; OT the polar subtangent ; 
and OD, the perpendicular from the pole on the tangent, is 
usually symbolized by p. The value of each of these lines is 
required. 

tan RQP = g = ^, from (1). (3) 

Since OPT = OQT + eld, the two angles OPT and OQT 
differ from each other by an infinitesimal, and therefore 
OPT = OQT, and hence, 

rrld 
tan OPT = ~ :> from (3), (4) 

sin OPT = sin OQP = ^ = ~, from (1). (5) 
Hence, 
OT = polar subtangent = OP tan OPT = -^ , 

from (4). (6) 
OC = polar subnormal = OP tan OPC = OP cot OPT 

= g, from (4). (?) 



PT = polar tangent = VoP 2 + OT 2 = r \/ \ + r 2 ^, 

a?' 2 

from (6). (8) 



PC = polar normal = VOP 2 + OC 2 = y r 2 + 



dr* 

d&' 

from (7). (9) 



180 



EXAMPLES. 



r 2 dd r 2 d0 

SD = p = OP sin OPD = ~ from (5) = 



Vr 2 dd 2 +dr 2 > 
from (2). (10) 



r See Price's Calculus, Vol. I, p. 417. 



EXAMPLES 



1. The spiral of Archimedes, whose equation is r = a9. 
(Anal. Geom., Art. 160.) 



Here 



fid 1 r 2 

^- = -; .'. Subt. = -, from (6). 

dr a a 

Subn. = a, from (7). 
Tangent = r j/l + ^, from (8), 



Normal == V*' 2 + ci l , from (9). 

^•2 



P 



vV 3 + a 2 

2. The logarithmic spiral r = a { 
Art. 163.) 

Here -j- = ae log « = r log a ; 

Subt = -. = mr, 

log a 

(where m is the modulus of the system 
in which log a . = 1). 



from (10). 

(Anal. Geom., 



Subn. = 



m 



P 



Normal == 



mr 



y / 1+ i «* + i 




Fig. 20 



W* 



c/r 2_ 1^ i 

r 2 + -— = (r 2 + r 2 log 2 a)*. 



RECTILINEAR ASYMPTOTES. 181 

Tan. OPT = ~ = ^- ; 
dr log a 

which is a constant ; and therefore the curve cuts every 
radius-vector at the same angle, and hence it is called the 
Equiangular Spiral. 

If a = e, the Naperian base, we have, 

tan OPT = .-!- = 1, and .-. OPT = 45°, 
loge 

and OT = OP = r. 

3. Find the subtangent, subnormal, and perpendicular in 

the Lemniscate of Bernouilli, r 2 = a 2 cos 20. (Anal. 
Geom., Art. 151.) 

Subtangent = -=—. — -— ; 
° a 2 sm 20 

a 2 
Subnormal = — - sin 2d ; 
r 

Perpendicular 



V/ 4 + ff*sin 2 20 « 2 

4. Find the subtangent and subnormal in the hyper- 
bolic* spiral rd = a. (Anal. Geom., Art. 161.) 

Subt. = — a ; Subn. = 

a 



RECTILINEAR ASYMPTOTES. 

103. A Rectilinear Asymptote is a line which is 
continually approaching a curve and becomes tangent to it 
at an infinite distance from the origin, and yet passes 
within a finite distance of the origin. < 

To find whether a proposed curve has an asymptote, we 
must first ascertain if it lias infinite branches, since if it 



* This curve took its name from the analogy between its equation and that of 
the hyperbola xy = a. (See Strong's Calculus, p. 145; also Young's Dif. Calculus, 
p. 120.) 



182 RECTILINEAR ASY3IPT0TES. 

has not, there can be no asymptote. If it has an infinite 
branch, we must then ascertain if the intercept on either 
of the axes is finite. The equation of the tangent 
(Art. 100) being, 

/ dy' v , N 

y -y = &*-*">* 

if we make successively y = 0, x — 0, we shall find for 
the intercepts on the axes of x and y, the following : 

dx 
Xo _x-y--, 

(by putting x = x and y = y , and dropping accents), 

dy 

y° = y- x tx' 

Now, if for x = oo both x and y are finite, they will 
determine two points, one on each axis, through which an 
asymptote passes. If for x = oo , x Q is finite and y infi- 
nite, the asymptote is parallel to the axis of y. If for 
x = go , x is infinite and ?/ finite, the asymptote is parallel 
to the axis of x. If both x Q and y are infinite, the curve 
has no asymptotes corresponding to x = oo. If both x 
and y are 0, the asymptote passes through the origin, and 

dy 

its direction is obtained by evaluating ~ for x = oo . 

When there are asymptotes parallel to the axis, they may 
usually be detected by inspection, as it is only necessary to 
ascertain what values of x will make y = oo, and what 
values of y will make x = oo . For example, in the equa- 
tion xy — m, x = makes y = oo, and y = makes 
# = oo ; hence the tAvo axes are asymptotes. Also in the 
equation xy — ay — bx = 0, which may be put in either 
of the two forms, 

bx ay 

y = - or x = — 2-^= ; 

x — a y — b 

y = oo when x = a, and a; •= oo when y = b ; 



EXAMPLES. 183 

hence the two lines x = a and y = b are asymptotes to 
the curve. 

In the logarithmic curve y = a x , 

y = when x = — oo , 

therefore the axis of x is an asymptote to the branch in the 

second angle. 

x z 

Also in the Cissoid y 2 = - , 

* 2a — x' 

y = oo when x = 2a ' K 
hence x = 2a is an asymptote. 

EXAMPLES*. 

1. Examine the hyperbola 

a 2 y 2 — b 2 x 2 = — a 2 b 2 , for asymptotes. 

Here 

dy b 2 x a 2 y 2 a 2 . . 

-f- = -r- ; .•. x = x — -jg- == - = for x = ± oo . 

y = y s— = = f or y = ± oo . 

J0 J a 2 y y 

Hence the hyperbola has two asymptotes passing through 
the origin. 

Ai dy b 2 x b 1 , I , 

Also -—- = -5 - = 4- = + - f or x = oo . 

ax a 2 y a / „2 ~~ a 

Hence the asymptotes make with the axis of x an angle 
whose tangent is ± - ; that is, they are the produced 
diagonals of the rectangle of the axes. 

2. Examine the parabola y % == 2p# for asymptotes. 



184 ASYMPTOTES DETERMINED BY EXPANSION. 

Here 

chi p y 2 , 

~ = - : .'. # = — 27- = — oo when x — y = co. 

dx y 2p u 

y = ^ = oo when y = oo or # = oo . 

Hence the parabola has no asymptotes. 
The ellipse and circle have no real asymptotes, since 
neither has an infinite branch. 

3. Examine y B = ax 2 4- x z for asymptotes. 

When x=±ac,y=±cc; .-. the curve has two 
infinite branches, one in the first and one in the third 
angle. 

dy _ 2ax -f dx 2 ■ 



a 

Xn — X - : ^— a — — ; 7T- .; — 77 > 



Vo = 



dx 


3f 


? 






of 






ax 2 




2ax + 3^ 


2ax + 3a; 2 




when x = 


00 . 






2ax 2 + 3a: 3 


3 


(r 


-X s ) - 


■ 2ax 2 



ty 2 3f 

ax 2 a , 

= ~, when x = oo 



3 («Z 2 + iB 3 )* 3 

Hence the asymptote cuts the axis of x at a distance 

— -, and that of y at a distance - from the origin, and as 
o o 

it is therefore inclined at an angle of 45° to the axis of x, 

its equation is 

a 

(See Gregory's Examples, p. 153.) 

104. Asymptotes Determined by Expansion. — A 

very convenient method of examining for asymptotes con- 
sists in expanding the equation into a series in descending 



EXAMPLES. 185 

powers of x, by the binomial theorem, or by Maclaurin's 
theorem, or by division or some other method. 

EXAMPLES. 



1. Examine y 2 = - for asymptotes. 

X ~~~ Qj 




Then 




, /x 4- a ,„ a a 2 , x 
y = ±x\/ — — = -j- a (1 + - + - 2 + etc.) 
J f x — a x x x 2 ' 


(1) 


When x = oo (1) becomes 




y — ± (x + «). 


00 



We see that as x increases, the ordinate of (1) diminishes, 
and when x becomes infinitely great, the difference between 
the ordinate of (1) and that of (2) becomes infinitesimal; 
that is, the curve (1) is approaching the line (2)' and 
becomes tangent to it when x = oo ; therefore, y = ± ( xJ t a ) 
are the equations of two asymptotes to the curve (1) at 
right angles to each other. 

Another asymptote parallel to the axis of y is given by 

x = a. 

2. Examine x 3 — xy 2 -f ay 2 — for asymptotes. 



Here ?/ = 4- \ / — - — 

J — V x ^ a 

/„ a Sa 2 off 3 , \ 

Hence, y = ± Ix + -j are the equations of the two 
asymptotes. 

By inspection, we find that x = a is a third asymptote. 

r 2 — 1 

3. Examine y 2 = x 2 — — - for asymptotes. 

X -4— X 



186 EXAMPLES. 

Here y = ±x\l 2 -f etc.j 

•'• V = ± # are the two asymptotes. 

105. Asymptotes in Polar Co-ordinates.— When 

the curve is referred to polar co-ordinates, there will be an 
asymptote whenever the subtangent is finite for x = oc . 
Its position also will be fixed, since it will be parallel to the 
radius- vector. Hence, to examine for asymptotes, we find 
what finite values of make r = oo ; if the corresponding 

flf) 

polar subtangent, r 2 -=-, which in this case becomes the 

perpendicular on the tangent from the pole, is finite or zero, 
there will be an asymptote parallel to the radius-vector. If 
for r = oo the subtangent is oo , there is no corresponding 
asymptote. 

EXAMPLES. 

1. Find the asymptotes of the hyperbola a 2 y 2 — b 2 x 2 = 
— a 2 b 2 by the polar method. 

The polar equation is 

n 2 h 2 
a 2 s'm 2 d-b i cos 2 d = --f- (1) 

r 2 

b 2 
When r = oo , (1) becomes, tan 2 d = — 2 ; 

.-. « = tan -.(±|). 

Therefore the asymptotes are inclined to the initial line 



at tan" 



(4Y 



From (1) we get ^ = _____^ , 

and r z d l _ + o»(Poo*g-*Bin'0)» (8) 

ana dr~ ± (a 2 + V) sin 6 cos 6 ' y ' 



EXAMPLES. 187 

which is equal to when = tan~M± -); "hence both 
asymptotes pass through the pole. 

2. Find the asymptotes to the hyperbolic spiral rS = a. 
(See Anal. Geom.", Art. 161.) 

Here r = -, .-. r = <x> , when 6 — 0. 



r 


= 


a 
6' 


dd 
dr 


= 


— 



a 



dd 



- 2 , and f»-= -a. 



There is an asymptote therefore which passes at a distance 
a from the pole and is parallel to the initial line. 

3. Find the asymptotes to the lituus rd* = a. (Anal. 
Geom., Art. 162.) 

Here r = — , .-. r = oo, when = 0. 
6? 

dO _ 2a? and r2 d6 = _ 2adl _ when e _ ^ 

Therefore the initial line is an asymptote to the lituus. 

4. Find the asymptotes of the Conchoid of Nicomedes, 
r = j3 sec + wr. (Anal. Geom., Art. 151.) 

• Here r = oc when = Jr ; and, r 2 -y- = ;j when 6 = -• 

Therefore the asymptote cuts the initial line at right 
angles, and at a distance p from the pole. 



EXAMPLES. 

1. Find the equation of the tangent to 3# 2 — 2^—10 = 0, 
at x = 4. Ans. y=± .7127a; ± .8909. 

X s 

2. Find the equation of the tangent to if = , at 

x = 2. 



y = 2x — 2 and i/ = — 2x -f 2. 



188 EXAMPLES. 

3. Find the equation of the tangent to the Naperian 
logarithmic curve. Ans. y = y' (x — x' -f 1). 

4. At what point on y = x 2 — 3x 2 — 24=x -f 85 is the 

tangent parallel to the axis of x ? 

dy' 
[Here we must put -f~, = 0. See Art. 56a.] 

At (4, 5) and (-2, 113). 

5. At what point on y 2 = 2x B does the tangent make 
with the axis of x an angle whose tangent is 3, and where is 
it perpendicular? At (2, 4) ; at infinity. 

6. At what angle does the line y — \x + 1 cut the curve 
y 2 = Ax ? [Find the point of intersection and the tangent 
to the curve at this point; then find the angle betw r een this 
tangent and the given line.] 10° 14' and 33° 4'. 

7. At what angle does y 2 = lOx cut x 2 -f y 2 = 144 ? 

71° 0' 58". 

8. Show that the equation of a perpendicular from the 
focus of the common parabola upon the tangent is 

y = - - (* - 1p). 

9. Show that the length of the perpendicular from the 
focus of an hyperbola to the asymptote is equal to the semi- 
conjugate axis. 

10. Find the abscissa of the point on the curve 

y(x — l)(x — 2) = x — 3 
at which a tangent is parallel to the axis of x. 

x = 3 ± V%. 

11. Find the abscissa of the point on the curve 

y 5 — (x — a) 2 (x — c) 

at which a tangent is parallel to the axis of x. 

%c + a 

X = r 



EXAMPLES. 189 



xi 
12. Find the subtangent of the curve y 



Ans. 



V2a — x 
x (2a — x) 















da — 


- X 


13. 


Find the 


subtangent 


of the 


curve 


y 3 - 


-3axy+i 

2axy - 
ay - 


c 3 =0. 

-X 2 


14. 


Find the 


subtangent 


of the 


curve 


xy 2 


= a 2 (a - 
2(ax- 


-x). 



a 

15. Find the subnormal of the curve if = 2a 2 log x. 

a 2 
x 

16. Find the subnormal of the curve day 2 + « 3 = 2x*. 

x 2 
a 

x 3 

17. Find the subtangent of the curve y 2 = — - — -■ 

2x (a — x) 
~3a — 2x" 

18. Find the subtangent of the curve 

x 2 y 2 — (a + x) 2 (& - &). 

x\a + x) (ff - x 2 ) 
x 3 + ab 2 

19. Find the subnormal, subtangent, normal, and tangent 
in the Catenary 





y = W +e ~j 




Subnormal 


c / - _-\ 

= -(e c — e c ) ; normal 


c 






t 




-& -<?)*' 


(f- 



190 EXAMPLES. 

20. Find the perpendicular from the pole on the tangent 
in the lituus rdi = a. _ 2a 2 r 

P ~~ (r 4 + 4fl 4 )^ 

21. At what angle does y 2 = 2ax cut x"—3axy-\-y 3 = ? 

cot -1 v 4. 

22. Examine y 2 = 2x + 3x 2 for asymptotes. 

y = V&e-I — -— is an asymptote. 

23. Examine y 3 = 6x 2 + x z for asymptotes. 

y = a; -f 2 is an asymptote. 

24. Find the asymptotes of # 2 (x — 2d) == a; 3 — « 3 . 

a? = 2# ; y = ± (sc + a). 

25. Find the asymptotes of y = ~~ ^ +JL. 

J * 5 z 2 — 3&Z + 2& 2 

a; = 5 ; a; = 23 ; y = x — 3 (a — b). 



CHAPTER X. 

DIRECTION OF CURVATURE, SINGLE POINTS, 
TRACING OF CURVES. 



106. Concavity and Convexity. — The terms concav- 
ity and convexity are used in mathematics in their ordinary 
sense. A curve at a point is concave towards the axis of x 
when in passing the point it lies between the tangent and 
the axis. See Fig. 21. It is convex towards the axis of x 
when its tangent lies between it and the axis. See 
Fig. 22. 

If a curve is concave down- 
wards, as in Fig. 21, it is plain 
that as x increases, a decreases, 
and hence tan a decreases ; that 

is, as x increases, ~ (Art. 56a) 
decreases ; and therefore the de- 
rivative of -ir or -~ is negative. 
dx dx* ° 

In the same way if the curve is 
convex downward, see Fig. 22, it is 
plain that as x increases, a in- 
creases, and therefore tan a in- 
creases ; that is, as x increases, 




Fig. 21. 



(Art. 12.) 



dx 



increases, and therefore the de- 



rivative of ~ or 
dx 



dx 2 




is positive. 



Fig. 22 



Hence the curve is concave or convex downward according 

dhj . 
as _ 1S _ or +. 



192 



POLAR CO-ORDINATES. 



This is also evident from Fig. 23, where MM' = M'M" 
= dx ; PP' is common to the two curves and the common 
tangent. PE = PR' == dx ; and PR 
= P 2 R'. But P"R' > P 2 R' > P,R'. 
Now P'R and P"R' are ' consecutive 
values of dy in the upper curve, 
and P'R and P,R' are consecutive 
values of dy in the lower curve, and 
hence P"R' - P'R = d (dy) = d 2 y is 
+ , and P,R' — P'R = d 2 y is — ; that 
is, d 2 y is — or -{-, according as the 
curve is concave or convex downwards. 
<Py 
dx} 
since dx 2 is always positive. 

We have supposed in the figures that the curve is above 
the axis oix. If it be below the axis of x, the rule just 
given still holds, as the student mav show by a course of 
reasoning similar to the above. 




The sign of 



is of course the same as that of 



dhi 
If the curve is concave downwards, -—{ is 

dx 2 

dhj 



if it be 



above the axis of x, y is + ; therefore, y -~ is — ; if the 



dhi 
curve be concave upwards, -~ is -f ; if it be below the 

axis of x, y is — ; therefore, y -~ is — ; that is, y —^ 
when the curve is concave towards the axis of x. In 



is 



dh! 



the same way it may be shown that y -— is + , when the 
curve is convex towards the axis of x. 



107. Polar Co-ordinates. — A curve referred to polar 
co-ordinates is said to be concave or convex to the pole at 
any point, according as the curve in the neighborhood of 
that point does or does not lie on the same side of the tan- 
gent as the pole. 



EXAMPLES. 



193 



It is evident from Fig. 24, that when the curve is con- 
cave toward the pole 0, as r increases p increases also, and 

therefore -=- is positive ; and if the curve is convex toward 
dp 

the pole, as r increases p decreases, and 

therefore V is negative. If therefore 
dp 

the equation of the curve is given in 

terms of r and 6, to find whether the 

curve is concave or convex towards the 

pole, we must transform the equation 

into its equivalent between r and p, by 

means of (10) in Art. 102, and then find 



Fig. 24. 




EXAMPLES. 

1. Find the direction of curvature of 



Here 



&y _ _2 

dx> ~ (x — %y 



d 2 ii 
that is, y| is positive or negative, according as z< or >2 ; 

and therefore the curve is convex downward for all values 
of x < 2, and concave downwards for all values of x > 2. 

2. Find the direction of curvature of 

y = h -j- c (x -f- a) 2 and y = a 2 V% — a. 

Arts. The first is concave upward, the second is concave 
towards the axis of x. 



3. Find the direction of curvature of the lituus r = 



*■ 



194 SINGULAR POINTS. 



dr r r z 
Mere dd~ ~W~ ~2tf> '** 


dr 2 r 6 

W 2 ~ I^ ; 


tiich in (10) of Art. 102 gives, 




2a 2 r dr 


(4a 4 4- r 4 )^ 


V " ( r 4 + ^4)1-' dp~ 


2a 2 (4a 4 — r 4 ) 



Therefore the curve is concave toward the pole for values 
of r < a V% and convex for r > a a/2. 

4. Find the direction of curvature of the logarithmic 
spiral r = a 6 . 

By Art. 102/ Ex. 2, 

mr dr \/m 2 4- 1 

P = 



Vm 2 + 1 d P 



m 



which is always positive, and therefore the curve is always 
concave toward the pole. 

SINGULAR POINTS. 

108. Singular Points of a curve are those points 
which have some property peculiar to the curve itself, and 
not depending on the position of the co-ordinate axes. 
Such points are : 1st, Points of maxima and minima ordi- 
nates ; 2d, Points of inflexion ; 3d, Multiple Points ; 4th, 
Cusps ; 5th, Conjugate points ; 6th, Stop points ; 7th, 
Shooting points. We shall not consider any examples of 
the first kind of points, as they have already been illus- 
trated in Chapter VIII, but will examine very briefly the 
others. 

109. Points of Inflexion. — A point of inflexion is a 

point at which the curve is changing from convexity to 

concavity, or the reverse ; or it may be defined as the 

point at which the curve cuts the tangent at that point. 

d 2 y 
When the curve is convex downwards, j~ is + (Art. 



EXAMPLES. 195 

106), and when concave downwards, -y^ is — ; therefore, 

d 2 y 
at a point of inflexion -j~ is changing from -f to — , or 

from — to + , and hence it must be or oo . Hence to 

dhi 
find a point of inflexion, we must equate j~ to or go , 

and find the values of x ; then substitute for x a value a 
little greater, and one a little less than the critical value ; 

if -j~ changes sign, this is a point of inflexion. 

cix 



EX A M PL E S . 

1. Examine y = b -f- (x — a) z for points of inflexion. 
Here g = 6(*-a)=0; 

.-. x = a and hence y = b. 

This is a critical point, L e., one to be examined ; for if 
there is a point of inflexion it is at x = a. For x > a, 

-=^ 2 is +, and for x < a, -~ 2 is — . Hence there is a point 

of inflexion at (a, l). 

2. Examine the witch of Agnesi, 

x 2 y = 4# 2 (2a — y), 
for points of inflexion. 

There are points of inflexion at ( ± — -, — )• 

3. Examine y = b + (x — «) _1 for points of inflexion. 

There is a point of inflexion at (a, b). 

4. Examine the lituus for points of inflexion. 

By Art. 107, Ex. 3, -j- is changing sign from + to — 
when r = a V~, indicating that the lituus changes at this 



196 METHOD OF FINDING MULTIPLE POINTS. 

point from concavity to convexity, and hence there is a 
point of inflexion at r = a V& 

110. Multiple Points. — A multiple point is a point 
through which two or more branches of a curve pass. If 
two branches meet at the same point, it is called a double 
point ; if three, a triple point ; and so on. There are two 
kinds : 1st, a point where two or more branches intersect, 
their several tangents at that point being inclined to each 
other ; and 2d, a point where two or more branches are 
tangent to each other. The latter are sometimes called 
points of osculation. 

As each branch of the curve has its tangent, there will 
be at a multiple point as many tangents, and therefore as 

many values of ~- as there are branches which meet in 

this point. If these branches are all tangent, the values of 

-~ will be equal. At a multiple point y will have but one 

value, while at points near it, it will have two or more 
values for each value of x. In functions of a simple form, 
such a point can generally be determined by inspection. 
After finding a value of x for which y has but one value, 
and on both sides of which it has two or more values, form 

~- If this has unequal values, the branches of the curve 

intersect at this point, and the point is of the first kind. If 

cly 

-=- has but one value, the branches are tangent to each 

cix 

other at this point, and the point is of the second kind. 

When the critical points are not readily found by inspec- 
tion, we proceed as follows: 

Let f(x 9 y) = (1) 

be the equation of the locus freed from radicals. Then 



EXAMPLES. 



197 



dy _ 
dx 



du 
dx 

dy 



and as differentiation never introduces radicals when they 
do not exist in the expression differentiated, the value of 

-f- cannot contain radicals, and therefore cannot have sev- 
ax 

eral values, unless by taking the form 







Hence we have 



dy 

dx 



du 

° r dx 



du 



0, and — = 0, from 

which to determine critical values of x and y. If these 

values of x and ?/ found from — — and -=- = are real 
J dx dy 

and satisfy (1), they may belong to a multiple point. If y 

has but one value for the corresponding value of x, and on 

both sides of it y has two or more real values, this point is a 

multiple point. We then evaluate -# = - , and if there 



are several real and unequal values of ~ , there will be as 



dx ~ ' 

dy 

dx' 

many intersecting branches of the curve passing through 
the point examined. (See Courtenay, p. 190.) 



EXAMPLES. 

1. Determine whether the curve y = (x — a)Vx 4- b has 
a multiple point. 

Here // has two values for every 
positive value of x > or < a. When 
x = or a, y has but one value, b; 
hence there are two points to be ex- 
amined. When x < 0, y is imagi- 
nary ; hence the branches do not 
pass through the point (0, b), and 




198 EXAMPLES. 

therefore it is not a multiple point. When x > or < a, y 
has two real values, and therefore (a, b) is a double point, 

dy , 3x — a ,- , 

Therefore the point is of the first kind, and the tangents 
to the curve at the point make with the axis of x angles 
whose tangents are -f V 'a, and — ^/a. 

2. Examine x* -f 2ax 2 y — ay z = for multiple points. 

We proceed according to the second method, as all the 
critical points in this example are not easily found by inspec- 
tion. 

du 



-^ = ix (3? + ay) = ; 


(i) 


^ = a(2x*-3f) = Q; 


(2) 


dy 4;e 3 + 4=axy 
dx ~ Say 2 — 2ax 2 


(3) 



Solving (1) and (2) for x and y, we find 

ix = 0\ lx — ia<\/6\ (x = —iaV$\ 
\t/ = o) ; \y= -ia) ; \y= -%a )' 

Only the first pair will satisfy the equa- 
tion of the curve, and therefore the ori- 
gin is the only point to be examined. 

du 
Evaluating — in (3) for x = and 

y = 0, and representing d £ hj P> and J 
by p', for shortness, we have 




CUSPS. 



199 



dy 

dx 



= P 



4# 3 + iaxy _ 
%wf~— &azj ~ ' 

12a; 2 + bay + 4a:rp 

Go?//; — 4aa: 
24^ + 8ap + 4o#j!7 
6a/; 2 + Qayp' — 4a 

6ap 2 -^4a' 

.*. p (6ap 2 



when 



P = °) 

\y = 0/ 







= g. when C = 3 



^ = 






— 4a) = 8a/? 
0, + V%, or 



when 



V& 



/* = 0\ 

\y = 0/ 



Hence the origin is a triple point, the brandies being in- 
clined to the axis of x at the angles 0, tan -1 (a/2), and 
tan -1 ( — V%), respectively, as in the figure. (See Courte- 
nay's Calculus, p. 191 ; or Young's Calculus, p. 151.) 

3. Examine y 2 — x 2 (1 — x 2 ) = for multiple points. 
Ans. There is a double point at the origin, the branches 

being inclined to the axis of % at angles of 45° and 135° 
respectively. 

4. Show that ay 2 —x s y—ax z = has no multiple points. 

111. Cusps. — A cusp is a point of a curve at which two 
branches meet a common tangent, and 
stop at that point. If the two branches 
lie on opposite sides of the common tan- 
gent, the cusp is said to be of the first 
species ; if on the same side, the cusp is 
said to be of the second species: 

Since a cusp is really a multiple point 
of the second kind, the only difference 
being that the branches stop at the point, 
instead of running through it, we exam- 
ine for cusps as we do for multiple points ; and to distin- 





Fig. 27. 



200 cusps. 

guish a cusp from an ordinary multiple point, we trace the 

curve in the vicinity of the point and see if y is real on one 

side and imaginary on the other. To ascertain the kind of 

cusp, we compare the ordinates of the curve, near the point, 

with the corresponding ordinate of the tangent ; or ascertain 

the direction of curvature by means of the second derivative. 

In the particular case in which the common tangent to 

the two branches is perpendicular to the axis of x, it is best 

to consider y as the independent variable, and find the 

, j, ax , 
values oi -=- , etc. 
dy 

EXAMPLES. 

1. Examine y = x 2 ± %% for cusps. 

We see that when x = 0, y has but one value, 0; when 
x < 0, y is imaginary ; and when x > 0, y has two real 
values; hence, (0, 0) is the point to be examined. 

dy 

dx 
a common tangent to both branches, and 
there is a cusp at the origin. 

-~ = 2 ±-^2^ is positive when x = 0; 
ax 

hence the cusp is of the first kind. 

dhi 
The value of ~ shows that the upper 

branch is always concave upward, while the lower branch 
has a point of inflexion, when x — -gfe ; from the origin to 
the point of inflexion this branch is concave upward, after 
which it is concave downward. 

The value of -^- shows that the branch is horizontal 
dx 

when x = ^f . From y = x?- — X s *, we find that the lower 

branch cuts the axis of x at x = 1. The shape of the curve 

is given in Fig. 26. 



2x ± \x* = 0, when x = ; hence the axis of x is 





CONJUGATE POINTS. 201 

2. Examine (y — If = (a; — a) z for cusps. 

^?zs. The point (a, b) is a cusp of the first kind. 

3. Examine cy 2 = x* for cusps. 

The origin is a cusp of the first kind. 

112. Conjugate Points. — A conjugate point is an iso- 
lated point whose co-ordinates satisfy the equation of the 
curve, while the point itself is entirely detached from every 
other point of the curve. 

For example, in the equation y = (a + x)Vx, if x is 
negative, y is, in general, imaginary but for the particular 
value x = — a, y — 0. Hence, P is a 
point in the curve, and it is entirely 
detached from all others. When x = 0, 
y = 0, which shows that the curve 
passes through the origin. For positive 
values of x, there will be two real values Fig. 29. 

of y, numerically equal, with opposite 
signs. Hence, the curve has two infinite branches on the 
right, which are symmetrical with respect to the axis of x. 

If the 6rst derivative becomes imaginary for any real 
values of a; and y, the corresponding point will be conjugate, 
as the curve will then have no direction. It does not fol- 
low, however, that at a conjugate point ~- will be imagi- 
nary; for, if the curve y =if(x) have a conjugate point at 
(x, y), from the definition of a conjugate point, we shall 
have 

f(x ± h) = an imaginary quantity. But 

/(* ± ;*) = r± f xl + J g ± J 6 - + etc., 

therefore, if either one of the derivatives is imaginary, the 
first member is imaginary; hence, at a conjugate point 
some one or move of the derivatives is imaginary. 

Since at a conjugate point some of the derivatives are 
imaginary, let the n th derivative be the first that is imagi- 



202 EXAMPLES. 

nary. Suppose the equation of the curve to be freed from 
radicals, and denoted by u = f (x, y) = 0. Take the n th 
derived equation (Art. 88,-Sch.); we have 

du d n y d n u 

dx dx 11 ' ' ' ' + da? " ? 

where the terms omitted contain derivatives of u with re- 
spect to x and y, and derivatives of y with respect to x, of 

du 
lower orders than the n th . If, then, -=r be not 0, the value 

d"ii 
of -j~ii obtained from the derived equation will be real, 
cix 

du 
which is contrary to the hypothesis; hence, — = is a 

necessary condition for the existence of a conjugate point. 
But 

du du dy _ 

dx dy dx ~~ ' 

therefore, since -=- = 0, we must have -=- = 0. Hence, at 
dy dx 

a conjugate point we must have -y- = 0, and — = 0. 

Rem.— Owing to the labor of finding the higher derivatives, it is 
usually better, if the first derivative does not become imaginary, to 
substitute successively a + h and a — h for x, in the equation of the 
curve, where a is the value of x to be tested, and h is very small. If 
both values oiy prove imaginary, the point is a conjugate point. 



EXAMPLES. 

1. Examine ay 2 — x z + 4:Cix 2 — 5a 2 x + 2a s = for con- 
jugate points. 



du 

dx 

du 
dy 



— Sx 2 + Sax — oa 2 = 0. (1) 

2ay = 0. (2) 



SHOOTING POINTS. — STOP POINTS. 203 

Solving (1) and (2), we get 



G = 3 and C=ft 



Only the first pair of values satisfies the equation of the 
curve, and hence the point (a, 0) is to be examined. 

dy _ _3x 2 — 8x + 5« 2 _ 6x — 8a 



Tx=? = 


2ay "lap 




= 


, when ( ~~ ); 

p \y = 0/ 




p* = 


- 1 ; .: p = ± V^l 


dx 



therefore, 



This result being imaginary, the point (a, 0) is a conju- 
gate point. 

2. Show that ar 4 — ax 2 y — axif -f a 2 y 2 = has a conju- 
gate point at the origin. 

3. Examine (c*y — x z ) 2 = (x — a) n (x — b) G for conjugate 
points, in which ay b. 

The point ib, J is a conjugate point. 

The first and second derivatives are real in this example ; hence the 
better method of solving it will be to proceed according to the Remark 
above given 

113. Shooting Points are points at which two or more 
branches of a curve terminate, without having a common 

tangent. 

Stop Points are points in which a single branch of a 
curve suddenly stops. 

These two classes of singular points but rarely occur, and 
never in curves whose equations are of an algebraic form. 



204 



EXAMPLES. 



EXAMPLES. 
X 



1. Examine y = - L for shooting points. 



Here 



1 + 

dy _ 



x (l + e x ) 



-\2 



If x is + and small, y is -f ; if 

# is — and small, y is — . When 

# is -+- and approaches 0, y = 0, 



and 



cfo/ 



; when x is — and ap- 




Fig. 30. 



proaches 0, y = 0, and -~ == 1. 

Hence, at the origin there is a shooting point, one branch 
having the axis of x as its tangent, and the other inclined 
to the axis of x at an angle of 45°. (See Serret's Calcul 
Differentiel et Integral, p. 267.) 

2. Examine y = log x. 

When x is +, y has one real value ; when x = 0, y = ; 
when x < 0, ?/ is imaginary ; hence there is a stop point at 
-the origin. 

3. Examine y = x tan -1 -• 



If 





X 


x 2 + 1 


# = + or 


-o, 2/ 


clll 77 

-/ = 5 or 

«£ 2 


, 77 

~~ 2* 



0; 



Hence the origin is a shooting point, the tangent being 
inclined to the axis of x at angles tan -1 (1.5708) and 
tan" 1 (- 1.570.8). 

4. Show that y = e 7 ~ x has. a stop point at the origin. 



TEA CIXG CUE VES: 205 

114. Tracing Curves.— We shall conclude this chap- 
ter by a brief statement of the mode of tracing curves by 
means of their equations. 

The usual method of tracing curves consists in assigning 
a series of different values to one of the variables, and cal- 
culating the corresponding series of values of the other, thus 
determining a definite number of points on the curve. By 
drawing a curve or curves through these points, we are 
enabled to form a tolerably accurate idea of the shape of the 
curve. (See Anal. Geometry, Art. 21.) 

In the present Article we shall indicate briefly the man- 
ner of finding the general form of the curve, especially at 
such points as present any peculiarity, so that the mind can 
conceive the locus, or that it may be sketched without 
going through the details of substituting a series of values, 
as was referred to above. 

To trace a curve from its equation, the following steps 
will be found useful : 

(1.) If it be possible, solve it with respect to one of its 
variables, y for example, and observe whether the curve is 
symmetrical with respect to either axis. 

(-?.) Find the points in which the curve cuts the axes, 
also the limits and infinite branches. 

(3.) Find the positions of the asymptotes, if any, and at 
which side of an asymptote the corresponding branches lie. 

(Jf) Find the value of the first derivative, and thence 
deduce the maximum and minimum points of the curve, the 
angles at which the curve cuts the axes, and the multiple 
points, if any. 

(5.) Find the value of the second derivative, and thence 
the direction of the curvature of the" different branches, and 
the points of inflexion, if any. 

(6".) Determine the existence and nature of the singular 
points by the usual rules. 



206 EXAMPLES. 



EXAMPLES, 



1. Trace the curve y = — 



+ x* 



When x = 0, y = ; .*. the curve passes through the 
origin. 

For all positive values of x, y is positive ; and when 
x = oo , y = 0. For negative values of x, y is negative, and 
when x = — oo , y = ; hence the curve has two infinite 
branches, one in the first angle and one in the third, and the 
axis of x is an asymptote to both branches. 

ty - ! — ^ . fiy _ 2x ( x2 — 3 ) 

dx ~~ (1 + tff ' dx 2 ~ (1 + tf 2 ) 3 ' 

When ar = + 1, -^ = ; .*. there is a maximum ordinate 

at x — + 1, and a minimum ordinate at a; = — 1, at 
which points y == J and — J respectively. 

When a; = 0, -j- = 1 ; .*. the curve cuts the axis of a at 
an angle of 45°. 

Putting the second deriva- 
tive equal to 0, we get x = 
or ± V3. Therefore, there 
are points of inflexion at (0, 0) -_ 
and at x == + V3 and — a/3, 
for which we have y = J\/3, 
— }V3. From a; = — V3 
to a; = -1- V3, the curve is concave towards the axis of x, 
and beyond them it is convex. 

From this investigation the curve is readily constructed, 
and has the form given in the figure. 

2. Trace the curve y z = 2ax 2 — xK 

y =z x* (2a — x)* ; 





Fig. 31. 



EXAMPLES. 



207 



dx 



4:(IX — Sx 2 



dx * 9#3 (2a 



8« 2 



*y 



When x = or 2«, y = ; .-.the curve cuts the axis of 
x at the origin and at x == 2rt. 

To find the equation of the asymptote, we have 



V = 



(■-!)'= -•(»-£- )■• 



therefore, ?/ = — a: + f a is the equation of the asymptote, 
and as the next term of the expression is positive, the curve 
lies above the asymptote. 

Evaluating the first derivative for x = 0, y = 0, we have 

4a — 6z 



dy _ 4«a: — 3:z 2 
dx ~ 3f~~ 



% 



dx 



/dy\ 2 4a 

\dx) = ty = x ' When 

% = /25 _ 

<** \/ Sy~ 



y = 0; 



± oo , when 



Hence, at the origin there are 
two branches of the curve tangent 
to the axis of y\ and the value of 

dy a 



fZ# 




shows that if y be negative as it 

approaches 0, -£- will be imaginary ; 

and hence the origin is a cusp of 
the first species. 

When x — ±a, -f- — ; .-. there is a maximum ordinate 
, 3 dx 

at £ = fa. 

When x = 2a, -/ = — = oo.: .-. the curve cuts the axis 

of #, at the point x = 2a, at right angles. 



208 



EXAMPLES. 



Tatting the second derivative equal to oo, we get x = 2a. 
When x < 2a, the second derivative is — , and when > 2a 
it is + ; hence the left branch is everywhere concave down- 
ward, and the right branch is concave downward from x = 
to x = 2a. At this last point it cuts the axis of x at right 
angles, and changes its curvature to concave upward; the 
two branches touch the asymptote at x = + oo and — oo , 
respectively, i. e., they have a common asymptote. 

In the figure, OA = 2a, OB = §a, 00 = \a. 

2a\ 



3. Trace the curve y = x I ) 

y \x — a! 



Let 







/. y = 0. 
x < a ; .\ y is positive. 

x = a; y = oo. 

^ > a < 2a ; y is negative. 
x = 2a ; «/ = 0. 

# > 2$; ?/ is positive, 

a? = oo ; y= co. 

When a? is — , y is always negative. 

To find the asymptote, we have 

2a 
x 




Fig. 32.a 



y = x 



a 

x J 



-•( 



2a* 



— etc. ) = ^ — a € 



.♦. y = x — a is the equation of the asymptote. 

Hence, take OB = a = OD, and the line BD produced 
is the asymptote; also take 00 = 2a. Then, since y = 0, 
both, when a; = and x = 2a, the curve cuts the axis of x 



EXAMPLES. 209 

at and C. Between and B. the curve is above the axis ; 
at B the ordinate is infinite ; from B to C, the curve is 
below ; from C to infinity, it is above OX. Also, it* x is 
negative, y is negative ; therefore the branch on the left of 
is eutirely below the axis. 

dy x 2 — 2ax + 2a 2 



Also, 



dx (x — ay 



Let x = a ; .*. -f = <*> ; and the infinite ordinate at the 
dx 

distance a to the right of the origin is an asymptote. 

If x = 0, -- = 2; if x = 2a, ~ = 2; i. e., the curve 

cuts the axis of x at the origin and the distance 2a to the 
right, at the same angle, tan -1 (2). 

If x 2 — 2ax + 2a 2 or (x — a) 2 + a 2 = 0, x is impossible ; 
hence there is no maximum or minimum ordinate. 

. . cPi/ 2(x — a) 2 — 2\(x — a) 2 -\-a 2 ] 

Again, yi = -^ — - — ^— — '- i 

& dx 2 (x — af 

— 2a 2 



~ (x — af ' 

.*. -^ is -I- if x < a, and is — if x > a. 
ax* 

But x < a, y is + ; and x > a < 2a, y is — ; and 

# > 2a, y.is + \ therefore, from to B, and B to C, the 

curve is convex, and from C to infinity, it is concave to the 

axis of x. 

d 2 y 2 a 2 

If X be ~' dtf = (^"+^j i is> +, but y is - ; therefore 

the branch from the origin to the left is concave to the axis 
of x. (See Hall's Calculus, pp. 182, 183.) 

4. Trace the curve y 2 = < 

The curve passes through the origin : is symmetrical 
with respect to the axis of x ; has a cusp of the first kind at 



210 TRACING POLAR CURVES. 

the origin; both branches are tangent to the axis of x ; are 
convex towards it; are infinite in the direction of positive 
abscissas, and the curve has no asymptote or point of in- 
flexion. 

115. On Tracing Polar Curves. — Write the equation, 

if possible, in the form r =f(6) ; give to such values as 

to make r easily found, as for example, 0, Jtt, n, §7r, etc. 

civ 
Putting -rz = 0, we find the values of for which r is a 

maximum or minimum, i. e., where the radius of curvature 
is perpendicular to the curve. 

Find the asymptotes and direction of curvature, and 
points of inflexion. After this there will generally be but 
little difficulty in finding the form of the curve. 

EXAMPLES. 

1. Trace the lituus r = — • 

When = 0, v = oo ; when = 1 (= 57°.3),* r= ±a; 
when = 2 ( = 114°.6), r = ± .Ha ; when = 3, r — 

■±_ .58a, etc. ; when = oo , r = 0. 

civ V^ d/V 

-7^ = — — « , and when -= f7l = 0, v = ; hence, r and 

«0 2a 2 a0 

are decreasing functions of each other throughout all their 

values ; f and the curve starts from infinity, when = 0, 

and makes an infinite number of revolutions around the 

pole, cutting every radius- vector at an oblique angle, and 

reaching the pole only when = oo . . 

eld 2a? 
The subtangent v 2 -y- = = 0, when r = oo; hence 

the initial line is an asymptote (Art. 105). 

* The unit angle is that whose arc is equal to the radius, and is about 57°. 29578. 
t If we consider alone the branch generated by the positive radius-vector. 



EXAMPLES. 211 

% = 2a " (4 ^ ~ -^ (see Art. 107, Ex. 3) ; hence there is 
dr ( 4a 4 + r 4)f 

a point of inflexion at r = aV%; from r = to r = a^/2 
the curve is concave toward the pole, and from r = a\/2 to 
r = oo it is convex. 

2. Trace the curve r = a sin 30. 

r = 0, when (9 = 0, 60°, 120°, 180°, 240°, and 300°. 
When 6 = 2n, or upwards, the same series of values recur. 

If 6 = 30°, 90°, 150°, 210°, 270°, and 330°, r = a, — a, 
a, — a, a, and — a, successively. 

dr 

— = 3a cos 36, showing that r begins at when 6 = 0, 

increases till it is a when 6 — 30°, diminishes to as 6 

passes from 30° to 60°, continues to diminish and becomes 

— a w 7 hen 6 becomes 90°, and so on. 

dp 18a 2 r - 8r 3 ■ ■ 

~ = o, which show T s that 

dr ( 9fl 2 _ 8,2)1 

the curve is always concave towards the 
pole. There is no asymptote, as r is 
never oo . 

Hence the curve consists of three 

Fig. 33. 

equal loops arranged symmetrically 
around the pole, each loop being traced twice in each revo- 
lution of r. A little consideration will show that the form 
of the curve is that given in the figure. (See Gregory's 
Examples, p. 185 ; also Price's Calculus, Vol. I, p. 427.) 

3. Trace the Chordel r = a cosec (^-)- 

\2nJ 

If 6 = 0, nn, 2nn, 3mr, 4nn, hwrr, etc., successively, 

r = oo , a, oo , — a, — oo , a, etc. 

dr a 6 a 9 6 ( 6\ 

-jt = — ~- cosec — cot — = — cosec 2 — ( — cos ^- 1 , 
d6 2n 2n 2n 2?i 2n\ 2nr 

which is negative from = to 6 = nn, positive from 




212 



EXAMPLES. 



6 = 7i?r to = 3ii7T, negative from 6 = Sun to = onn, etc. 
Hence we see that r begins at qo when = 0; diminishes 
till it becomes a when 6 = nn; increases as passes from 
nn to 2nn ; becomes oc when = 2nn; when passes 2?in 9 
r changes from + go to — oo ; when increases from 2mr 
to 371TT, r increases from —go to — a; when increases 
from 3?iT7 to 4wtt, r diminishes from — a to — co ; when 
passes 4?ztt, r changes from — oo to + go . When in- 
creases beyond 4=n, the same values of r recur, showing that 
the curve is complete. 




Fig, 110. a 



dr a „ 

-^ = -- cosec 2 
dO 2n 



2n\ 



e 

2n> 



— ) = gives = ^tt, 3wrr, 



bnn, etc. ; i e., the radius- vector is a minimum at = W7r, 
3w7r, 57in, etc. 

<:?0 2w« 



The subtaugent 



r 2- _ 

dr 



and 



cos 







2w 

— 2na when = 0; 
+ 2wa when = 2wtt ; 



EXAMPLES. 213 

therefore the curve has two asymptotes parallel to the initial 
line, at the distances ± %na from the pole. 

r 2 _ 2anr 

P ' ~ 1^1 . A* " (4A 2 - a 2 + r 2 )* ' 

dp 2ahi (in 2 — 1) 



dr 



[a 2 (4w 2 - 1) + r 2 ]t 



3 5 



/. the curve is always concave towards the pole. 

Thus it appears that while is increasing from to 2nn, 
the positive end of the radius-vector traces the branch 
drawn in Fig. 34 ; and while d increases from 2nn to 4^rr ? 
the negative end of the radius-vector traces a second branch 
(not drawn), the two brandies being symmetrical with 
respect to the vertical line through the pole 0. 



EXAMPLES. 

1. Find the direction of curvature of the Witch of Agnesi 

x 2 y = ia 2 (2a — y). 

The curve is concave downward for all values of y between 
2a and -§«, and convex for all values of y between §# and 0. 

2. Find the direction of curvature of y = I? + (x — «) 3 . 

. Convex towards the axis of x from x > a to x = co ; and 
from x = a — b* to x = — oo ; concave towards the axis 
of x from x < a to x = a — b*. 

3. Examine y = (a — x)* -f ax for points of inflexion. 

There is a point of inflexion at x = a. 

4. Examine y .= x -f 36^ 2 — 2a; 3 — a 4 for points of in- 
flexion. Points of inflexion at 8 = 2, 8 = — 3. 



214 EXAMPLES. 

5. Find the co-ordinates of the point of inflexion of the 
curve 

X ? ( a 2 _ X 2) 

x = ± — ; y = jfea. 

V6 

«0 2 

6. Examine r = ^ z for points of inflexion. 

6r 1 

c?r 2 _ 4?' (r — a) 3 

Here ^ 7 a^~~ ; 

rtr 2 
• ^ — • • ptc 

(4r* — 12flr» + 13« 2 r 2 — 4aV)* ' 
There are points of inflexion at r = }a and r = -§#. 

7. Examine y 2 — (x — 1) 2 £ for multiple points. 

There is a multiple point at # = 1. 

3 ?(fl 2 — a 2 ) 

a 2 + a 2 

There is a multiple point at the origin, and the curve is 
composed of two loops, one on the right and the other on 
the left of the origin, the tangents bisecting the angles be- 
tween the axes of co-ordinates. 



8. Examine y 2 = — ^ % ^ or multiple points. 



9. Show that tf + x 2 y 2 — 6ax 2 y + a 2 y 2 = has a multiple 
point of the second kind at the origin. 

10. Show that y = a -f x -f- hx 2 ± ex* has a cusp of the 
second kind at the point (0, a), and that the equation of the 
tangent at the cusp is y = x + #. 

11. Show that y z = ax 2 + x z has a cusp of the first kind 
at the origin. 

12. Show that ay 2 — x z + bx 2 = has a conjugate point 

46 
at the origin, and a point of inflexion at x = -^ • 

o 



EXAMPLES. 215 

13. Trace the curve y % — a z — x 3 . 

The curve cuts the axes at {a, 0) and (0, a). 
It has an asymptote which passes through the origin. 
The points where the curve cuts the axes are points of 
inflexion. 

14. Trace the curve y = ax 2 ± Vbx sin x. 

For every positive value of x there are 
two values of y, and therefore two 
points, except when sin x = 0, in which 
case the two points reduce to one. 
These points form a series of loops like pig, 35, 

the links of a chain, and have for a 
diametral curve the parabola y = ax 2 , from which, when x 
is positive, the loops recede and approach, meeting the 
parabola whenever x = or re, or any multiple of rr. But 
when x is negative, y is imaginary except when sin z = 0, 
in which case y = az 2 , so that on the negative side there is 
an infinite number of conjugate points, each one on the 
parabola opposite a double point of the curve. (See De 
Morgan's Cal., p. 382 ; also, Price's Cal., Vol. I, p. 396.) 




CHAPTER XI. 



RADIUS OF CURVATURE, EVOLUTES AND INVO- 
LUTES, ENVELOPES. 

116. Curvature. — The curvature of a curve is its rate 
of deviation from a tangent, and is measured by the external 
angle between the tangents at the extremities of an indefi- 
nitely small arc ; that is, by the angle between any infini- 
tesimal element and the prolongation of the preceding 
element. This angle is called the angle of contingence of 
the arc. Of two curves, that which departs most rapidly 
from its tangent has the greatest curvature. In the same 
or in equal circles, the curvature is the same at every 
point ; but in unequal circles, the greater the radius the 
less the curvature ; that is, in different circles the curvature 
varies inversely as their radii. 

Whatever be the curvature at 
any point of a plane curve, it is 
clear that a circle may be found 
which has the same curvature as 
the curve at the given point, and 
this circle can be placed tangent 
to the curve at that point, with 
its radius coinciding in direction 
with the normal to the curve at 

the same point. This circle is called the osculating circle, 
or the circle of curvature of that point of the curve. The 
radius of curvature is the radius of the osculating circle. 
The centre of curvature is the centre of the osculating circle. 

For example, let ABA'B' be an ellipse. If different 
circles be passed through B with their centres on BB', it is 




Fig. 36. 



ORDER OF COX TACT OF CURVES. 



217 



clear that they will coincide with the ellipse in very differ- 
ent degrees, some falling within and others without. Now; 
that one which coincides with the ellipse the most nearly 
of all of them, as in this case MN, is the osculating circle 
of the ellipse at B, and is entirely exterior to the ellipse. 
The osculating circle at A or A', is entirely within the 
ellipse ; while at any other point, as P, it cuts the ellipse, 
as will be shown hereafter. 



117. Order of Contact of 
Curves. — Let y = / (x) and y 
= (x) be the equations of the 
two curves, AB and ab, referred to 
the axes OX and OY. Giving to 
x an infinitesimal increment h, and 
expanding by Taylor's theorem, we 
have, 




Fig. 37, 



y, == f(x + h) = f(x) +/' (x) h + /" (x) i 



W 



y, = (x + 7i) 



+ 4>" 0*0 s-« + ^c. 



(1) 



(2) 



Now if, when x = a = OM, we have f(a) ==. 6 (a), the 
two curves intersect at P, i. e., have one point in common. 
If in addition we have /' (a) = <p' (a), the curves have a 
common tangent at P, i. e., have two consecutive points in 
common ; in this case they are said to have a contact of the 
first order. If also we have, not only /(a) = (a) and/' (a) 
= <p ! (a), but/" (a) = $'' (a) ; i. e., in passing along one of 

d?ii 
the curves to the next consecutive point, j-£ (i. e., the curva- 
ture), remains the same in both curves, and the new point 
10 



218 CONTACT OF THE SECOND ORDER. 

is also a point of the second curve ; i. e., the curves have 
three consecutive points in common ; in this case the curves 
are said to have a contact of the second order. If f (a) 
= 4> (a),f (a) = 0' (a),f (a) = 0" {a) J'" (a) = 0'" (a), 
the contact is of the third order, and so on. It is plain 
that the higher the order of contact, the more nearly do 
the curves agree; if every term in (1) is equal to the cor- 
responding term in (2), .then y 1 = y 2 , and the two curves 
become coincident. 

118. The Order of Contact depends on the num- 
ber of Arbitrary Constants. — In order that a curve may 
have contact of the n th order with a given curve, it follows 
from Art. 117 that n + 1 equations must be satisfied. 
Hence, if the equation to a species of curve contains n + 1 
constants, we may by giving suitable values to those con- 
stants, find the particular curve of the species that has 
contact of the n th order with a given curve at a given point. 
For example, the general equation of the right line has two 
constants, and hence two conditions can be formed, / (x) 
= (x) and/' (x) = <p' (x), from which the values of the 
constants may be determined so as to find the particular 
right line which has contact of the first order with a given 
curve at a given point. In general, the right line cannot 
have contact of a higher order than the first. 

Contact of the second order requires three conditions, 
f(x) = (f>(x), f'(x) = 0'(a?), and f"(x) = 0" (x), and 
hence in order that a curve may have contact of the second 
order with a given curve, its equation must contain three 
constants, and so on. The general equation of the circle 
has three constants; hence, at any point of a curve a circle 
may be found which has contact of the second order with 
the curve at that point ; this circle is called the osculating 
circle or circle of curvature of that point ; in general, 
the circle cannot have contact of a higher order than 
the second. The parabola can have contact of the 



RADIUS OF CURVATURE. 219 

third order, and the ellipse and hyperbola of the 
fourth. 

In this discussion we have assumed that the given curve is of such 
nature as to allow of any order of contact. Of course the order of 
contact is limited as much by one of tbe curves as by the other. For 
example, if the given curve were a right line and the other a circle, 
the contact could not in general be above the first order, although the 
circle may have a contact of the second order with curves whose 
equations have at least three constants. Also, we have used the 
phrase in general, since exceptions occur at particular points, some of 
which will be noticed hereafter. 

119. To find the radius of curvature of a given 
curve at a given point, and the co-ordinates of the 
centre of curvature . 

Let the equation of the given curve be 

y =/(*). (i) 

and that of the required circle be 

(x' _ m f + $ _ n )% — r 2 . p) 

ifc is required to determine the values of m, n, and r. 

Since (2) has three arbitrary constants, we may impose 
three conditions, and determine the values of these con- 
stants that fulfil them, and the contact will be of the 
second order (Art. 118). 

From (2), by differentiating twice, we have, 



x' — m + <y 


-»)^ = 0; (3) 


* + ■£ + *■ 


— )g = a W 


If (2) is the circle of curvature at the point (x 3 y) of (1), 
3 must have, 


x' = X, 


y' = y\ 


<fy_dy 

dx' dz' 


rPy' _ d*y 

dx 2 ~~ dx 2 



220 RADIUS OF CURVATURE. 

Substituting these values in (2), (3), and (4), we have, 

(x _ m f + (y — nf = r 2 ; (5) 

x — m + (y — ri) ~ = (6) 



1 + ^ 2 + ^ w > dx 2 ~ 



\ ■*" dx 2 ! dx 



(7) 



u f i 2 

Therefore, y — n = -= (8) 

dx 2 

V ■*" dxV dx 

x - m = — § — < 9 > 

<fe 2 
By (5), (8), and (9), we have 

'•=±— ^— do) 

From (9) and (8) we have 



dx 2 

i + d -t 
n=zy ~W' ( } 

dx 2 

120. Second Method. — Let ds denote an infinitely 
small element of a curve at a pointy and the angle which 
the tangent at this point makes with the axis of x. Imagine 
two normals to be drawn at the extremities of this elemen- 
tary arc, i. e., at two consecutive points of the curve ; these 



RADIUS OF CURVATURE. 29A 

normals will generally meet at a finite distance. Let r be 
the distance from the curve to the point of intersection of 
these consecutive normals. Then the angle included be- 
tween these consecutive normals is equal to the correspond- 
ing angle of contingency (Art. 116), i. e., equal to cl(p. Since 
d<p is the arc between the two normals at the units distance 
of the point of intersection, we have 

ds = rd<p, or r = -j-> (1) 

Now this value of r evidently represents the radius of the 
circle, which has the same curvature as that of the given 
curve at the given point, and hence is the radius of curva- 
ture for the given point, while the centre of curvature may 
be defined as the point of intersection of two consecutive 
normals. 

To find the value of r, we have (Art. 56a), 

tan <f> = -4- ; .-. d = tan -1 -/ ; 

dx dx ' 

dx 



and hence d(j> = —=— 2 ; also, ds = Vdx 2 + dxf 

1 + %L 
+ dx* 



Substituting in (1), we have 



i + m 

cVy 
dx* 



W 



which is the same as (10) of Art. 119. 
As the expression (1 + -^-J" has always two values, the 

one positive and the other negative, while the curve can 
generally have only one definite circle of curvature at any 
point, it will be necessary to agree upon which sign is to be 



222 RADIUS OF CURVATURE. 

taken. We shall adopt the positive sign, and regard r as 
positive when the second derivative is positive, i. e., when 
the curve is convex downwards. (Usage is not uniform on 
this point. See Price's Calculus, Vol. I, p. 435. Todhun- 
ter's Calculus, p. 339, etc.) 

121. To Find the Radius of Curvature in Terms 
of Polar Co-ordinates. 

We may obtain this by transforming (2) of Art. 120 to 
polar co-ordinates, from which we find 



("*» 



n= NS 



where N is the normal. See Art. 102, Eq. 9. [See (2) of 
Ex. 4, Art. 90.] 

122. At a Point where the Radius of Curvature 
is a Maximum or a Minimum, the Circle of Curva- 
ture has Contact of the Third Order with the Curve. 

Since r is to be a maximum or a minimum, we must 

have -=2 = 0. 
dO 

Differentiating (2) of Art. 120 with respect to x, we have 

2V "*" dxV dx\dxV dxA ^ dxV 



dr 
dx 



= 0; 



3 dy/cW 
cfiy _ dx\dxV 
dtf~~ " % 3 ' 
+ ^ 2 



(1) 



DIFFERENT ORDERS OF CONTACT. 223 

Differentiating (8) of Art. 119, we have 

cPy _ dx\dz 2 ) . . 

dx 3 ~ dif ^ ' 

+ dtf 

Hence the third derivative at a point of maximum or 

minimum curvature is the same as it is in the circle of 

curvature, and therefore the contact at this point is of the 

third order (Art. 117). 

Cor. — The contact of the osculating circles at the 
vertices of the conic sections is closer than at other 
points. 

123. Contact of Different Orders. — Let y — f{x) 

and y- = fa) represent two curves, and let x } be the ab- 
scissa of a point of their intersection ; then we have 

Substituting x x ± h for x x in both equations, and sup- 
posing y x and y 3 the corresponding ordinates of the two 
curves, we have 



ft 



f\x x ± h) = f{x x ) +/' (*) (± h) +r (^)- ( -# 



2 



+/' /> (^)%.|-^etc (1) 

f 9 = cp fa ± h) = fa) + 0' fa) (± h) + 0" (*,) ^^ 

-f 0"' (*) ^=^-° + etc. (2) 

Subtracting (2) from (1), we get, for the difference of 
their ordinates, corresponding to x ± It, 

y> - y> = L/"(*0-* '(*)] (±*) + [/"(*.) - *"(*)] ^- 2 

+ [r , (*.)-*'"(*.)] i | r ^ • (3) 



224 EXAMPLES. 

Now if these curves have contact of (he first order, the 
first term of (3) reduces to zero (Art. 117). If they have 
contact of the second order, the first two terms reduce to 
zero. If they have contact of the third order, the first three 
terms reduce to zero, and so on. Hence, when the order of 
contact is odd, the first term of (3) that does not reduce to 
zero must contain an even power of ±- h, and y } — y 2 does 
not change sign with h, and therefore the curves do not 
intersect, the one lying entirely above the other ; but when 
the order of contact is even, the first term of (3) that does 
not reduce to zero must contain an odd power of ± h, and 
V\ — #2 changes sign with li, and therefore the curves inter- 
sect, the one lying alternately above and below the other. 

Cok. 1. — At a point of inflexion of a curve, the second 
derivative equals ; also, the second derivative of any point 
of a right line equals 0. Hence, at a point of inflexion, 
a rectilinear tangent to a curve has contact of the 
second order, and therefore intersects the curve. 

Cor. 2. — Since the circle of curvature has a contact of 
the second order with a curve, it follows that the circle of 
curvature, in general, cuts the curve as well as 
touches it. 

Cok. 3. — At the points of maximum and minimum curva- 
ture, as for example at any of the four vertices of an ellipse, 
the osculating circle does not cut the curve at its point of 
contact. 

EXAMPLES. 

1. Find the radius of curvature of an ellipse, 

t . t - 1 

a* "*" & ~ 

_ dy Wx , df aY + Wx* 

Here -f ■• = «- ; .*. 1 -f ~r~h — A <> — * 

dx a 2 y' dx 2 aY 





EXAMPLES. 221 


<Py _ 


aWy — aWx-f 

u dx _ b 2 (a 2 y 2 + fix 2 ) b* 


dx 2 


a^y 2 a 4 y* d l y z 



.-. (Art. 120), r 



dx 2 



= ^h^XT (neglecting the sign). 



b 2 



At the extremity of the major axis 

x = a, y = 0, .-. r = - 

At the extremity of the minor axis, 

x = 0, y = o, .: r = y 

2. Find the radius of curvature of the common parabola, 



y 2 = 2px. 



Here dy_P ^y__v 2 . 

dx~y' dx 2 ~ if 

0/ + j0 2 ) ¥ __ (normal) 3 
p 2 p 2 

At the vertex, y = ; . •. r = p. 

3. Find the radius of curvature of the cycloid 



x = r vers -1 - — V2r^/ — # 2 . 

TT dx y _ dip 2r 

Here -- = — -_-=^=\ .-. 1 + ~h = — 5 
<ty ^/2ry — f c ^ V 

which equals twice the normal (Art. 101, Ex. 5). 



226 



E VOLUTES AND INVOLUTES. 



4. Find the radius of curvature of the parabola whose 
latus-rectum is 9, at x = 3, and the co-ordinates of the 
centre of curvature. r = 16.04; m = 13-|-, n = — - 6.91. 

5. Find the radius of curvature of the ellipse whose axes 
are 8 and 4, at x = 2, and the co-ordinates of the centre of 
curvature. r = 5.86; m = .38, n = — 3.9. 

6. Find the radius of curvature of the logarithmic spiral 

r = a e . 



dr 

dd 



R 



= a 6 log a ; 
(r 2 -f- r 2 log 2 a) ■ 



5P 



a 9 log 2 a ; 



( r 2 + T % i og 2 fl j| _ jv; 



r 2 + 2r 2 log 2 a — r 2 log 2 a 
(See Ex. 2, Art. 102.) 

7. Find the radius of curvature of the spiral of Archi- 

es _i_ r 2)h 
medes, r = aO. E = -WV 2 * 

2« 2 4- r 2 

8. Find the radius of curvature of the hyperbolic spiral, 

r (a 2 + r 2 )i 

a 3 



rd — a. 



E 



124. Evolutes and Involutes. — The curve which is 
the locus of the centres of all the osculating circles of a 
given curve, is called the evolute of that curve ; the latter 
curve is called the involute of the former. 

Let P„ P 2 , P 3 , etc., represent a series of 
consecutive points on the curve MN, and 1 
C„ C 2 , C 8 , etc., the corresponding centres 
of curvature ; then the curve C 1? 2 , C 3 , 
etc., is the evolute of MN, and MN is the 
involute of G u C 2 , C 3 , etc. Also, since the 
lines CjPi, 0. 2 P 2 , etc., are normals to the 
involute at the consecutive points, the 
points d, C 2 , C 3 , etc., may be regarded as 




c < Fig. 38. 



EQUATION OF THE E VOLUTE. 227 

consecutive points of the evolute ; and since each of the 
normals PA, P 2 C 2 , etc., passes through two consecutive 
points on the evolute, they are tangents to it. 

Let r„ r 2 , r z , etc., denote the lengths of the radii of 
curvature at P„ P 2 , P 3 , etc., and we have, 

r 2 -r 1 = P 2 C 2 - PA = PA - PA = C A ; 

also r z - n = P3C3 - PA = P 8 C, - PA = C A ; 

and i\ — r z == C 3 C 4 , and so on to r n ; 

hence by addition we have, 

r n - r, = CA + CA + C„_A, 

This result holds when the number n is increased indefi- 
nitely, and we infer that the length of any arc of the 
evolute is equal, in general, to the difference between 
the radii of curvature at its extremities. 

It is evident that the involute may be generated from its 
evolute by winding a string round the evolute, holding it 
tight, and then unwinding it, each point in the string 
will describe a different involute. It is from this property 
that the names evolute and involute are given. While a 
curve can only have one evolute, it can have an infinite 
number of involutes. 

The involutes described by two different points in the 
moving string, are said to be parallel ; each curve being got 
from the other by cutting off a constant length on its 
normal, measured from the involute. (Williamson's Dif- 
ferential Calculus, p. 295.) 

125. To find the Equation of the Evolute of a 
Given Curve. — The co-ordinates of the centre of curva- 
ture are the co-ordinates of the evolute (Art. 124). Hence, 
if we combine (11) and (12) of Art. 119 with the equation 
of the curve, and eliminate x and y, there will result an 
equation expressing a relation between m and n, the co-or- 



228 



EXAMPLES. 



dinates of the required eve-lute, which is therefore the 
required equation ; the method can be best illustrated by 
examples. 

The eliminations are often quite difficult ; the following 
are comparatively simple examples. 



EXAMPLES 



1. Find the equation of the evolute of the parabola, 

y 2 = 2px. (1) 



Here 



% _P . &y 



f 



dx y ' dx 2 y z 

Substituting in (11) and (12) of Art. 
119, we have, 

V 2 + V % V V s 



m 



x = 



y r 

m — p 



y 2 + p 2 y B 




# 9 



p* p* 

.-. y = —phi*. 
And these values of x and y in (1) give, 
p*n* — \p (m — p) ; 



»2 — 



27^ 



(m — pf ; 



(») 



which is the equation required, and is called the semi-cubical 
parabola. Tracing the curve, we find its form as given in 
Fig. 39, where AG = p. 

If we transfer the origin from O to A, (2) becomes 

2 8 3 



EXAMPLES. 



229 



2. Find the length of the e volute AQ , Fig. 39, in terms 
of the co-ordinates of its extremities. 

Let ON = x, NQ = y ; ON' = m, N'Q' = n. 
Then by Art. 123, Ex. 2, we have, 
(if + tf)l 



r = 



^ 



Therefore, by Art. 124, we have, 
Length of AQ' = Q Q - AO = ^ + ffi -p 

= (»^ + J$)^ — J*. (Since y 2 = p%n* 9 by Ex. 1.) 

3. Find the equation of the evolute of the cycloid, 



r vers" 



v*ry - y 2 - 



(i) 



Here 



dy _ v%ry — ?/ 2 d 2 y 
dx ~ y ' dx 2 



or 



m = x + 2 ^*lry — ;/ 2 and » == — y ; 

a; = m - 2 V- 2''^ — w 2 and y = — n; 



which, in the equation of the cycloid, gives 

( n \ y 
;) + V —2r)i—ii 2 ,(2) 

which is the equation of a cycloid 
equal to the given cycloid ; the 
origin being at the highest point, o 
This will appear by transforming 

x = *r versm -1 - -j- \Z%ry — ?/ 2 , (3) 




230 PROPERTIES OF EVOLUTES. 

(which gives points between 0' and X), to parallel axes 
through 0'. 

Denoting by m and n the new co-ordinates, the formulae 
for transforming from to 0' are 

x = nr + m, y = 2r + n ; 

which in (3) gives 

nr + m = r vers -1 (2 + -) + V%r(2r + n) — (2r + n) 2 
= r vers -1 2 + r vers -1 ( ; j + V— 2rn — n 2 ; 

.-. m = r vers -1 ( ) + V— 2m — n 2 ; (4) 

which is the same as equation (2). Hence we see that the 
equation of the evolute, OA (2), is the same as that of the 
cycloid, O'X (4). That is, the evolute of a cycloid is an 
equal cycloid.* 

126. A Normal to an Involute is Tangent to the 

Evolute.— This was shown geometrically in Art. 124; tt 
may also be shown as follows : 

Let (x, y) be any point P of the involute (Fig. 40), from 
which the normal PQ is drawn, and let {m, n) be the point 
Q on the evolute through which the normal passes. 

The equation of PQ is 

y-n= - C -^(x-m); (1) 

or x - m +-£(y~ n) = 0. (2) 

Now when we pass from a point P to a consecutive 
point on the involute, Q also will change to a consecutive 



This property was first discovered by Huygens. 



ENVELOPES OF CURVES. 231 

point of the evolute, therefore we differentiate (2) with 
respect to x, regarding x, y, m, n, as variables, and get 

1 dx + aW {V n) + da* dx dx~ a 

But, since (m, ^) is the centre of curvature corresponding 
to P, we have, by (8) of Art. 119, 






which in (3) gives 



dm dy dn _ dx _ dn m 

dx dxdx ~ dy ~~ dm 9 

and this in (1) gives 

y — n = c j^ (* - m ) ; (*) 

therefore (1) or (4), which is the equation of a normal to 
the involute at P (x, y), is also the equation of a tangent to 
the evolute at Q (ni, n). 

127. Envelopes of Curves. — Let us suppose that in 
the equation of any plane curve of the form 

f{x,y,a) = 0, (1) 

we assign to the arbitrary constant a, a series of different 
values, then for each value of a we get a distinct curve, 
different from any of the others in form and position, and 
(1) may be regarded as representing an indefinite number 
of curves, each of which is determined when the correspond- 
ing value of a is known, and varies as a varies. 

The quantity a is called a variable parameter, the name 
being applied to a quantity which is constant for any one 
curve of a series, but varies in changing from one curve to 
another, and the equation f(x, y, a) = 0, is said to repre- 
sent & family of curves. 

If we suppose a to change continuously, i. e., by 



232 EQUATION OF THE ENVELOPE. 

infinitesimal increments, the curves of the series represented 
by (1) will differ in position by infinitesimal amounts ; and 
any two adjacent curves of the series will, in general, inter- 
sect; the intersections of these curves are points in the 
envelope. Hence an envelope may be defined as the locus 
of the intersection of consecutive curves of a series. 

It can be easily seen that the envelope is tangent to each 
of the intersecting curves of the series ; for, if we consider 
four consecutive curves, and suppose Pj to be the point of 
intersection of the first and second, P 2 that of the second 
and third, and P 3 that of the third and fourth, the line 
Pi P 2 joins two infinitely near points on the envelope and 
on the second of the four curves, and hence is a tangent 
both to the envelope and the second curve ; in the same 
way it may be shown that the line P 2 P 3 is a tangent to the 
envelope and the third consecutive curve, and so on. 

128. To Find the Equation of the Envelope of a 
given Series of Curves. 

Let f{x, y, a) = 0, (1) 

f(x, y, a + da) = 0, (2) 

be the equations of two consecutive curves of the series; 
then the co-ordinates of the points of intersection of (1) a"nd 
(2) will satisfy both (1) and (2), and therefore also will 
satisfy the equation 

• A *' y ' a) ~i ( ' T,y ' a+tfa) = ° < Anal - aeom -> Art 3 °)> 

&&*4 = 0, (3) 

da v ' 

and therefore the points of intersection of two infinitely 
near curves of the series satisfy each of the equations (1) 
and (3). Hence, to find the equation of the envelope, we 
eliminate a between (1) and (3), i. e., we eliminate the varia- 
ble parameter between the equation of the locus and its first 
differential equation. 



EXAMPLES. 



233 



EXAMPLES 



m 



1. Find the envelope of y = ax -\ , when a varies. 

Differentiating with respect to a, x, and y, being constant, 
we have 



a 4 \xl 

y "= ± \j^mx + Vmx] or ^/ 2 = 4*MaJ a 



which is the equation of a parabola. 

2. A right line of given length 
slides down between two rectangular 
axes ; to find the envelope of the hue 
in all positions. 

Let c be the length of the line, a 
and b the intercepts OA and OB ; 
then the equation of the line is 



x y 



(1) 




Fig. 41. 



in which the variable parameters a and. b are connected by 
the equation 

tfi + b 2 = c 2 . (2) 

Differentiating (1) and (2), regarding a and b as varia- 
ble, we have 



— ~ a da — %- n db = 0, or — — . da = %- n db. 



ada + bdb = 0, 1 
Dividing (3) by (4), we have 



or 



y 

l_ 



a = (.rc 2 )3 and # 



ada 



a" 1 " b 
& + V 



bdb. 



(3) 



■8' 



234 



EXAMPLES. 



which iii (2) gives 

2 2 2 

.*. 0?3 + ?/3 = c», 

which is the equation required. 
The form of the locus is given in 
Fig. 42, and is called a hypo-cycloid, 
which is a curve generated by a 
point in the circumference of a 
circle as it rolls on the concave 
arc of a fixed circle. 




Fig. 42. 



3. Find the envelope of a series of ellipses whose axes are 
coincident in direction, their product being constant. 



Here 






- + £ - 1,' 

« 2 + 5 2 




(1) 


Let 






a • ' J = c ; 




(2) 


•*• 


X % 

a* 


da -f 


^-^ = 0, or '-da = 
b 3 a° 


-f>- 


(3) 






da 
a 


db _ rffl 

-f — = 0, or — = 
b a 


db 
b ' 


(4) 



Dividing (3) by (4), Ave have 



x II 

- =z -j~ — i, by substituting in (1). 



a = x\/2, 
from (2), 



and 



b = yV2; 



xy = 



r 



which is the equation of an hyperbola referred to its asymp- 
totes as axes. 

This example may also be solved as follows : Eliminating 
b from (1) and (2). we have 



EXAMPLES. 235 



JS + $*■ = !. TO 



in which we have only the variable parameter a. 



a y n 2 cx /r\ 



which in (5) gives 



4. Find the envelope of the right lines whose general 
equation is 

y = 7??^ + (aW -f # 2 )^, (1) 

where m is the variable parameter. 

"We find m = — , 

tt Vcc 2 — X 2 

X 2 I/ 2 

which in (1) gives — a + j- 2 = 1 for the required envelope. 

Hence the envelope of (1 ) is an ellipse, as we might have 
inferred, since (1) is a tangent to an ellipse. (See Anal. 
Geom., Art. 74.) 



EXAMPLES. 

1. Find the radius of curvature of the logarithmic curve 

X = log y. _ ( m 2 + y2)| 

my 

2. Find the radius of curvature of the cubical parabola 
f = & x > ^ _ {9>if + a*f 

T ~ 6a*y 

3. Find the radius of curvature of the curve 

y = X s — x 2 + 1 



236 EXAMPLES. 

where it cuts the axis of y, and also at the point of mini- 
mum ordinate. 

At the first point, r = — -| ; at the second, r = |-. 

4. Find the radius of curvature of the curve 

y 3 — Qx 2 + x z . 

- [y 4 + (4a; + a?)*]* 
* ~ -~8xhf~ 

5. Find the radius of curvature of the rectangular hyper- 
bola xy = m 2 . _ ($ + y 2)i 

T ~ ~2m 2 

6. Find the radius of curvature of the Lemniscate of 
Bernouilli r 2 = a 2 cos 26. a 2 

7. Find the equation of the evolute of the ellipse 

a 2 y 2 + b 2 x 2 == aW. 

(fl«i)t + (&w)* = (a 2 - 6 2 )i 

8. Find the equation of the evolute of the hyperbola 

a 2 y 2 — b 2 x 2 = — a 2 !) 2 . 

(arn)% — (bn)% = (a 2 + b 2 )*. 

9. Prove that, in Fig. 39, OM = 40A = 4p, and MP' 

10. Find the length of the evolute AP' in Fig. 39. 

Ans. (3* — l)p. 

11. Find the length of the evolute of the ellipse. (See 

Art. 123, Ex. 1, and Art. 124.) 4 . a s — ¥ 

Ans. 4 — -. 

ao 

12. Find the length of the cycloidal arc OCX, Fig. 40. 

Ans. 8r. 

13. Find the envelope of the series of parabolas whose 
equation is y 2 = m (x — m), m being the variable parameter. 

■ x 

y=± r 



EXAMPLES. 237 

14. Find Jie envelope of the series of parabolas expressed 

by the equation y — ax ^ x 2 , where a is the variable 

parameter. 

The result is a parabola whose equation is 

a*=2p(f-y). 

This is the equation of the curve touched by the parabolas de- 
scribed by projectiles discharged from a given point with a constant 
velocity, but at different inclinations to the horizon. The problem 
was the first of the kind proposed, and was solved by John Bernouilli. 
but not by any general method. 

15. Find the envelope of the hvpothenuse of a right- 
angled triangle of constant area c. c 

*y = r 

16. One angle of a triangle is fixed in position, find the 
envelope of the opposite side when the area is constant = c. 

c 

*y = % 

17. Find the envelope of x cos a + y sin « = p, in 
which cc is the variable parameter. x 2 -f y 2 = p\ 

18. Find the envelope of the consecutive normals to the 
parabola if = 2px. 

o 

Ans. y 2 = 5=- (x — p) 3 , which is the same as was found 

a ip 

for the evolute in Ex. 1, Art. 125, as it clearly should be. 
(See Art. 124.) 

19. Find the envelope of the consecutive normals to the 
ellipse a 2 y 2 + b 2 x 2 = a?b 2 . 

Ans. (az)i + {by)* == (« 2 — b 2 )i, which is the same as 
was found in (?) for the evolute of the ellipse. 



PART II. 

INTEGRAL CALCULUS. 



• ♦ >-- 



CHAPTER I. 

ELEMENTARY FORMS OF INTEGRATION. 

129. Definitions. — The Integral Calculus is the inverse 
of the Differential Calculus, its object being to find the 
relations between finite values of variables from given 
relations between the infinitesimal elements of those vari- 
ables ; or, it may be defined as the process of finding the 
function from which any given differential may have been 
obtained. The function which being differentiated pro- 
duces the given differential, is called the integral of the 
differential The process by which we obtain the integral 
function from its differential is called integration. 

The primary problem of the Integral Calculus is to effect 
the summation of a certain infinite series of infinitesimals, 
and hence the letter S was placed before the differential to 
show that its sum w T as to be taken. This was elongated 

into the symbol / (a long S), which is the sign of integra- 
tion, and when placed before a differential, denotes that its 
integral is to be taken. Thus, / 3xMx, which is read, " the 

integral of d&dx," denotes that the integral of Sx 2 clx is to 
be taken. The signs of integration and differentiation are 



ELEMENTARY RULES FOR INTEGRATION. 239 

inverse operations, and when placed before a quantity, 
neutralize each other. Thus, 

/ d (ax) = ax, 
and d I axdx = axdx. 

130. Elementary Rules for Integration. — In the ele- 
mentary forms of integration, the rules- and methods are 
obtained by reversing the corresponding rules for differ- 
entiation. When a differential is given for integration, if 
we cannot see by inspection what function, being differ- 
entiated, produces it, or if it cannot he integrated by known 
rules, we proceed to transform the differential into an 
equivalent expression of known form, whose integral we 
can see by inspection, or can obtain by known rules. In 
every case, a sufficient reason that one function is the 
integral of another is that the former, being differentiated, 
gives the latter.* 

(i.) Since 

d (v -f y — • z) = dv + dy — dz ; (Art. 14.) 

/ (dv -f dy — dz) = Id (v + y — z) = v -j- y — z 

== / dv -f- I dy — dz. 

Hence, the integral of the algebraic sum, of any 
number of differentials is equal to the algebraic sum 
of their integrals. 

(2.) Since 

d (ax + b) = aclc ; (Art. 15.) 



* While there is no quantity whose differential cannot he found, there is a large 
cla^s of differentials whose integrals cannot be obtained ; either because there is no 
quantity which, being differentiated, will give thorn, or because the method for 
•their integration has not yet been found. 



240 ELEMENTARY RULES FOR INTEGRATION. 

I adx = J d {ax -f b) = ax ± b 
= a I dx ± b. 

Hence, a constant factor can he moved from one 
side of the integral sign to the other without affect- 
ing the value of the integral. Also, since constant 
terms, connected by the sign +, disappear in differentia- 
tion, therefore in* returning from the differential to the 
integral, an arbitrary constant, as C, must be added, whose 
value must be determined afterwards by the data of the 
problem, as will be explained hereafter. 

(3.) Since 

d ~ [/Ml" = a [/Ml"- 1 df(x) ; (Arts. 15 and 19.) 

.-. fa LA*)]- 1 df(x) = fd\ [/(*}]" = I [/(*)]■ + c. 

Hence, whenever a differential is the product of 
three factors, viz, a constant factor, a variable factor 
with any exponent except — 1, and a differential 
factor which is the differential of the variable factor 
without its exponent, its integral is the product of 
the constant factor by the variable factor with its 
exponent increased by 1, divided by the new ex- 
ponent* 

It will be seen that the rule fails when n = — 1, since 
if we divide by 1 — 1 = 0, the result will be oo . 

flflT 

(Jj..) Since d (a log x) = — ; (Art. 20, Cor.) 
•'• J~V ~ J d ^ l0g X) = a l0g X ' 



* The arbitrary constant is not mentioned since its addition is always under- 
stood, and in the following integrals it will he omitted, as it can always be supplied 
when necessary. 



EXAMPLES. 241 

Hence, whenever a differential is a fraction whose 
numerator is the product of a constant by the differ? 
ential of the denominator, its integral is the product 
of the constant by the JVapsrlau logarithm of the. 
denominator. 

EXAMPLES. 

1. To integrate dy = ax 5 dx. 

y — j axhlx — J a • x 5 • dx = — • [by (5)]. 

i. To integrate dy = (a -f- h&fjfidx. 

The differential of the quantity within the parenthesis 
being \hx % dx, we write 

y = ff, (a + o^ lfaftb = (g + £W . [by (3)1 

This example might also be integrated by expanding the quantity 
within the parenthesis, and integrating each term separately by (1), 
but the process would be more lengthy than the one employed. 

3. To integrate 

dy — a (ax 2 + fee 3 )* 2xdx + Zbx 2 (aa* + bxrf dx. 

y = C[ a (ax 2 + bx$)% 2xdx + Sbx 2 (ax 2 + &c»)* dx] 
= J (ax 2 + te*)i (2«z + 3^ 2 ) efo = | (ax 2 + fee 3 )! [by (J)]. 

4. To integrate % = 



a + to 

Since the numerator must be bdx to be the differential of 
the denominator, we must multiply it by b, taking care to 
divide by b also ; hence, 

p adx a p bdx a, , . z O ru / #\i 

V = I r- = t / t- =s t log (a + to), [by (4)1. 

^ J a + ox bJ a + bx b c v ; L J N /J 

11 





n+1 










a 






(n- 


- 1) x n - 


-i 


= 


a log x. 


* 





242 DIFFERENTIAL FORMS. 

131. Fundamental Forms. — On referring to the forms 
of differentials established in Chap. II, we may write down 
at once the following integrals from inspection, the arbitrary 
constant being aliuays understood. 

C „ . ax n+1 

1. y = I ax" ax 

Padx 

2. y — / 

Padx 

4. y =p J a x log a dx = ax. 

5. y = / e^z = e®. 

6. y = I cos # (?# s= sin #. 

7. ?/ = / — sin x dx = cos x. 

8. ?/ = / se ° 2 ^ ^ == ^ an x - 

9. y = I — cosec 2 a; $z = cot a% 

10. y = I sec a* tan x dx = sec a?. 

11. y z= I — cosec x cotana dx == cosec a;. 

12. y = I sin x dx = vers a;. 

13. y = I — cos a: t/a; = covers a;. 



* Since the constant c to be added is arbitrary, log c is arbitrary, and we may 
write the integral in the form 



/• 



— = a log x + log c = log ex? 
x 



INTEGRATION BY TRANSFORMATION. 24; 



1. r Clx 

14. y = / — = sin -1 x. 

lo. y ■=■ I = cos -1 x. 






16. y — J j—~~ = taE _1 ^ 



. y = /; — - = cot x a;. 

^ «/ r 1 + x 2 



dx 




\'Zx — x 2 



= sec -1 x. 
= cosec -1 re. 
= vers -1 x. 
= covers -1 x. 



These integrals are called the fundamental or elementary 
forms, to which all other forms, that admit of integration 
in a finite number of terms, can be ultimately reduced. It 
is in this algebraic reduction that the chief difficulty of the 
Integral Calculus is found ; and the processes of the whole 
subject are little else than a succession of transformations 
and artifices by which this reduction may be effected. The 
student must commit these fundamental forms to memory; 
they are as essential in integration as the multiplication 
table is in arithmetic. 

132. Integration of other Circular and Trigono- 
metric Functions by Transformation into the Fun- 
damental Forms. 

dx 

1. To integrate dy = — : 

Va' — J 8 ^ 

We see that this has the general form of the differential 



244 INTEGRATION BY TRANSFORMATION. 

of an arc in terms of its sine (see form 14 of Art. 131) ; 
hence we transform our expression into this form, as folio ws : 

1 



P dx P dx r a 



dx 



Va k — b 2 x 2 " b 2 x 2 " I Shy 



V a* \ a 



To make this quantity the differential of an arc in terms 
of its sine, the numerator must be the differential of the 
square root of the second term in the denominator, which 

is - dx. Therefore we need to multiply the numerator by b, 

which can be done by multiplying also by the reciprocal of 
b, or putting the reciprocal of b outside the sign of integra- 
tion. Hence, 

p dx _ r ~a dx _ IP a d 



/, b 2 x 2 W yk# 



1 . .bx 

■=■ sm _1 — • 
o a 



2. To integrate dy = 



dx 



Va 2 - b 2 x 2 
dx p dx 



Here y = / — — 

«/ „ /„2 7,2^.2 «/ 



^a 2 — b 2 x 2 J A b 2 x* 



A b 2 w 



lf-^4 



\/ 



hdx 1 .lx 

COS" 



If- 



b 2 x 2 b a 



a 



P dx 1 . .bx 

3. y = I — — — = — tan -1 — • 

* J a 2 + b 2 x 2 ab a 



-/■ 



dx 1 , .bx 

= -; cot -1 



a 2 -f b 2 x 2 ab a 



PROCESS OF INTEGRATION. 245 



'• y = f: 



dx 1 « bx 



Vb*x 2 



= - S8C" 



a a 



dx 1 . bx 



n dx 1 

6. y = / , = - cosee 



xy/bW—a? a' " J a 



y= I -7 tt — TT: = % vers_1 



f* dx 1 __, 5a; 

fe 1 „ bx 



Vtabx — b 2 x* b 

P dx 1 ^bx 

8. y = / ; — = y covers"" 1 — 

J V'2abx — Jm % a 

A /*, , Psmxdx pdcosx 

9. y = / tan x dx = / = — / 

9 *J J COS 2 «/ 



COS SB «/ COS # 

— log cos a; = log sec a;. 



/* , Pcosxdx 

10. y = I cot x dx = I — : = log sin a?. 

9 J J sin # 

J ~~ J sin a; -. »/ 2 sin Ja; cos Ja; 

/i sec 2 (ia;) ^ . 

12. , = /A. = f 

v J COS X J 



tan £a; 
<2ai 

cos x ~ 



= logtang + ^) [by (11)]. 

., P dx /^sec 2 a; tfz . 

13. y — I = / — r = log tan re, 

* J sin a; cos a; «/ tan a; 

/* rfa /* (sin 2 a; + cos 2 a*) dx 

~~ «/ sin 2 a; cos 2 a; t/ sin 2 a; cos 2 x 

(since sin 2 a; + cos 2 a; = 1) == / (sec 2 a; -f cosec 2 a;)^a; 



= tan x — cot x. 



15. V — J tan 2 x dx = J (sec 2 x — 1) dx = tan * " — & 



246 EXAMPLES. 

16. V — J cot# 2 dx = I (cosec 2 x — l)dx 

= — cot X — X. 

17. V = J cos 2 # dx = J (J 4- J. cos %x) dx (by Trig.) 

= \x + J sin 2a;. 
(See Price's Calculus, Vol. II, p. 68.) 

18. y = / sin 2 # efa; = J# — J sin 2x. 

Remark. — It will be observed that in every case we reduce the 
function to a known form, and then pass to the integral by simple 
inspection or by the elementary rules. Whenever there is any doubt 
as to whether the integral found be correct or not, it is well to differen- 
tiate it, and see if it gives the proposed differential. (See Art. 130.) 

EXAMPLES. 

1. dy = bx^ dx. 

Here y = / bx^dx = / b'X^-dz 

= §&»£ [by (3) of Art. 130]. 

_ , XCIX 

2. dy = — 

Here y = Jv^Tx^ = /* ^ 2 + ^' Hxdx 

= (a 2 + z 2 )i 

3. dy = %x%dx. y = fyN". 

4. dy = 2z$dx. y = £aA 

5. % = — ar&cfc, y = 5a;~i 

6. dy = -— • # = 2Vs. 



INTEGRATING FACTOR. 247 

7. dy = — bmx~*dx. y = — j /i?/p. 

8. dy = (f «a$ — -p.T*) tfe. # = ajrf — &#*. 

O 7 I 12 S \ 7 12 . X 

io. ^ = ^i^-^_^. 

(3 ax 2 — X s )* 

Here y = / — (3ax 2 — a; 3 )~ 3 (2#£ — x 2 ) dx 

= — I (3 as; 2 — a^)T. 

11. dy = (2ax — x 2 )^ (a — x) dx. y = \ (2ax — a?)*. 

133. Integrating Factor. — It has been easy, in the 
examples already given, to find the factor necessary to make 
the differential factor the differential of the variable factor 
[see (3) of Art. 130] ; but sometimes this factor is not easily 
found, and often no such factor exists. There is a general 
method of ascertaining whether there is such a factor or 
not, and when there is, of finding it, which will be given 
in the two following examples : 

12. dy = 3 (Abx 2 — 2cx^ (Ux — 3cx 2 ) dx. 

Suppose A to be the constant factor required ; then we 
have 

y = f^- (Ux 2 - 2cx s )s (12 Abx — SAcx 2 ) dx. 

If A be the required factor, we must have 

d (Abx 2 — 2cx*) = (12 Abx — 9Acx 2 ) dx, 

or Sbx — 6cx 2 = 12 Abx — 9Acx 2 , 

and since this result is to be true for every value of x, the 
coefficients of the like powers of x must be equal to each 



248 EXAMPLES. 

other, from the principle of indeterminate coefficients, giv- 
ing us two conditions, 

12Ab = 8b, (1) 

and — 9Ac = — 6c, (2) 

or A = f . from (1), and A = \, from (2) ; 

hence § is the factor required, and we have 

y = A (±bx 2 — 2afift (8bx — Sex*) dx 
== | (Abx 2 — 2cx*)i 

13. % = (2aa — # 2 )-i (5a — #)$& 
Let ^4 be the required factor, then 

y = I —r (2ax — x 2 )% (5Aa — Ax) dx. 

,\ d (2ax — x 2 ) =2 (hAa — Ax) dx ; 

or 2a — 2x = 5,4a — J a;. 

.-. 2a = 5^a; (1) 

and — 2 = — J ; (2) 

or J. = f, from (1), and A =± 2, from (2). 

These values of ^4 being incompatible with each other, 
we infer that the differential cannot be integrated in this 
form. 

14. dy = (2b -f- Sax 2 — 5x z )~* (2ax — 5a; 2 ) dx. 

y = J (2b -+- 3«a; 2 — oa; 3 )3. 

15. dy = (Sax 2 + 4=bx*)* (2ax + 46a: 2 ) ri& 

y = \ (Sax 2 + 4fe 3 )i 

16. £&/ = flK^rfa? H ~- .• y = -^ -4- a& 

* 2^2 3 



EXAMPLES. 249 

17. dy = 5 (ox 2 — 2x)* xdx — (ox 2 — 2x)~* dx. 
Here y = I (5a:" 2 — 2^)» (5a; — 1) dx. 

y = | (5a: 2 - 2x)i 

134. A Differential may often be brought to the 

form required in (3) of Art. 130, by transposing a variable 
factor from the differential factor to the variable factor, 
or vice versa. 

2adx 



18. dy = 

x V2ax — x 2 

Here y = I (2ax — x 2 )~? 2ax~ 1 dx 

= f - (2ax~ 1 — i)4 (— 2ax~ 2 ) dx; 
. . y = -2 (2ax~ 1 - 1)* [by (3) of Art. 130] 



2 V2ax — x 2 



x 



19. dy — r y = — - « 

(2a# — a; 2 ) 2 a y 2«.r — # 2 

7 «.rffe a# 

20. dy = — — y = 



(2&e + x 2 )? b V2bx + ^ 



_ 7 «>fo 2« V3to + U 2 x 2 

21. e?y = r v = ^r-^ 

a; V3^ + 4c 2 ;*; 2 3 ^ 

22. % = yl+ l^L^i [g (4) f Art 130 -| 

/•J (« -\- bx -\- ex 2 ) . . oV 

23. ^ = ^^j. y = log (« + 2**). 



250 EXAMPLES. 

.. 7 5x?dx 

24. cly = 



3x*+ 7 
V = A lo g ( 3 * 4 + ?) = lo g (3^ 4 + ^ 

25 « ^ = Jfj y = log (* + »*. 

26 - ^ = o^L- y = iog 



8^ - M* * a ($a- 3x*)i 

27. cly = (b — x^fxidx. 

y = fft*t _ <L$i x i + T 6_^V- _ JL^¥. 

28. d^ = Jj*L. "y.= log(a+.to') li . 

29. dy = 2 log 3 2: — y = i log 4 #• 

30. rfy = « 2 * log atfa;. [See (4) of Art. 131.] 

Here y = J a 2 * log acjx — J \a& log ad (2x) = |«2x. 

31. % = 3a x2 log a^t^. 2/ = i axa - 

32. cly = ba&dx. \ 



* 3 log a 

33. tfy = 5e x dx. y — 5e x . 

34. cly = cos 3atfa\ [See (6) of Art. 131.] 

y = J sin 3a;. 

35. c??/ = cos (re 3 ) xdx. y — 1 sin (a: 2 ). 

36. c?^ = e sin x cos ar/a;. «/ = e sin * 

37. cly = — e™ sx sin attfo. # = e cosx . 

38. tfy == sin 2 (2x) cos (2a;) cZa;. 

Here y — A sin 2 (2a?) cos (2a;) J (2a;) = -J- sin 3 (2a;). 



EXAMPLES. 251 

39. dy = cos 3 (2x) sin (2x) dx. y — — \ cos 4 (2a:). 

40. dy = sec 2 (xf x 2 dx. y — i tan (x) s . 

41. % == sec (3x) tan (3z) da:, y ~ \ sec (3a;). 

42. dy = sin (ax) dx. y = - vers (ax). 

43. dy = tan o:do:. «/ = log sec a;. 

44. dy = sin sec 2 0d0. y = sec 0. 

45. dy = -====• [See (14) of Art. 131.] 

y = sin -1 (2a:). 
. 7 rado: a • -1 / o\ 

O^da: 



47. dy = 

a/2 — 4^ 

Here y — J 



%da 



x*ax 



V2 • Vl — 2^ 

a/2 • §#*<fc 



— / 



A/2 • f < 7 a/2 • a/1 - 2Z 3 



= */: 



dA/2a: 3 



a/1 - 2a: 3 
.•. y = -J sin -1 a/22 3 . 

-o 7 dx Sx 

48. dy = ^ ^ m - y = -J- sin" 1 — . 



V2 


- 9x 2 


— 


xdx 


A/2 


— S^ 4 




dx 



49. rhj=- p^°L . ^ = * C o S -i/V5 A, 

J A/2-5a^ J 2 a/5 \ 2 */ 



50. dy = j==-J V = % cos" 1 1, t 

A/ ax — x 2 \ V a, 



252 




EXAMPLES. 


51. 


Sdx 
d * f 4 + 'fUT 


y = -|-tan- 1 y 


52. 


x 2 dx 

g y = tt&- 


y = i tan -1 a?. 


53. 


, odx 

dy = t = 


= . # = vers -1 3x. 



54. e?y 



V^a; — 9a; 2 
2dx 



7 Vox* — 5 

Here y= f _ 2 ^_ = = ■*= /* 

J x Vo Via; 2 — 1 V5* 7 



V5V|^ 2 -1 V5 47 Vl^Vf^-l 
V5 



'*• 2/ = TTh sec-M^ V|). 



2<7a; - - 

55. dy = ' • y = V% eosec -1 (x Vf). 

V 5a; 4 — 2a; 2 

56 - ^ = 77^=1' # = -7= cosec" 1 (a; a/- 1 ?). 

V14o; 2 — 3 v 3 

V 4a; 3 — 9« 6 ^ 

58. «# = — • y = a covers -1 3a; 2 . 

* V6 - 9a; 2 * 

59. ^ = SfJ ^L^. ^ = 4 a/6 vers -1 (to*). 

V 2a? — 6a;f 

X — ■* sin a; 

60. <&/ = — — dx. , ?/ = log (a; + cos x). 

X ~\~ COb X 

xdx 

68 - *» = S~S'Xlx ix - y = log (a ' 3 - foS + 9 ^ 



EXAMPLES, 253 



Mx _ _3_ 

2 + ox 2 J ~~ yio 



63 - fy = a L R* ' V = ~1= tan" 1 x VI. 



64 . ^fcL±^* r = f<. + V5)i 

V rJ 

*- 7 (2a 2 + 4a; 2 ) as /-s— ; — 2 

6o. o# = ==±: — y — 2x va z + a; 2 . 

66. <7?/ = tan 2 re sec 3 zda?. y — i tan 3 a*. 

67. dy = ± ^ ?/ = logs — a 8 + j. 

_ (a; — 2) clx r 4 

a ya; V a; 

x^clx 

69. dfy = -5 -• y = -J.T 3 — x + tan -1 a;. 



rO. cly = 



+ 1 
dx 



1 + x + x' 



Here V — j i 

o 



= -- /"■ 



_2^ 
V3 



v8 i+ [" + »t-J 

= -?= tan" 1 (a; + \) A- 
a/3 V3 

tfa: 

71. % = « ^.-r^To- ? = tan-i (x - 1). 



2 — 2x -f a 

72. ay = 



a 2 + # 2 



y = 7 tan_1 1 + I log (a2 + ^ 



254 Th'TGOXOMETETC REDUCTION. 

135. Trigonometric Reduction. — A very slight ac- 
quaintance with Trigonometry will enable the student to 
solve the following examples easily. After a simple 
trigonometric reduction, the integrals are written out by 
inspection. 

1. dy = tan 3 xdx. 

Here y = J tan 3 xdx = I (sec 2 x — 1) tan xdx 

= / [sec 2 x tan xdx — tan xdx] 

= I tan 2 x — log sec x. [See (9) of Art. 132.] 

2. dy = tan 4 xdx. 

y = \ tan 3 x — tan x -f z. [(15) of Art. 132.] 

3. dy = tan 5 xdx. 

y = J tan 4 a; — -§- tan 2 x + log sec #. 

4. dy = cot 3 scdfc. y = — | cot 2 a; — log sin #. 

5. dy = cot 4 £$£. y = — J cot 3 sc -f cot a; + x. 

6. dy = cot 5 xdx. 

y = — \ cot 4 a; + \ cot 2 a; + log sin x. 

7. dy = sin 3 xdx. 

Here y = / \ (3 sin a; — sin 3a:) da; ; (from Jrig.), 

.-. y = -jig- cos 3a: — f cos x ; 
or y = \ cos 3 x — cos #.* 

8. d?/ = cos 3 xdx. y = sin x — -J sin 3 a:. 



* By employing different methods, we often obtain integrals of the same 
expression which are different in form, and which sometimes appear at first sight 
not to agree. On examination, however, they will always be found to differ only 
by some constant ; otherwise they could not have the same differential. 



EXAMPLES. 255 

9. dy = cos 4 xdx. 

Here V — J (cos 2 xydx = J{[l-\-2 cos 2x 

+ cos 2 2a?] dx [See (17) of Art. 132.] 

= i# -f i sin 2x + $ J cos 2 2oy7 (2a;) 

= }a? + J sin 2a; + J [> + J sin 4a;] ; [by (17) of Art, 132,1 
,\ y = -g^ sin 4a; + J sin 2a; + fa;. 

10. dy = sin 4 a;efa;. y = ^ sin 4a; — J sin 2a; + fa;. 

11. <?y = sin 5 xdx. y = — cos x 4- | cos 3 a; — ^ cos 5 a;. 

12. d# = cos 5 a;<#a;. # = sin x — f sin 3 x -f- \ sin 5 x. 

13. 6?y = sin 3 a; cos 3 xdx. 

y = yf^ cos 6a; — -g^ cos 2a;. (From Ex. 7.) 

14. dy = 5 — «/ = \ sec 2 a; -f- log tan x. 

* sm x cos 3 a; ^ ° 

15. dy = sin 3 a; cos 2 xdx. y = -^ cos 5 a; — i cos 3 a;. 

16. rfy = cos 5 a; sin 5 xdx. 

V = — ^|~ (i — i cos2 * + i cos4 *)• 

17. % = sin 6 x cos 3 a;^a; = (sin 6 x — sin 8 x) cos xdx. 

= sin 7 a; (^ — £ sin 2 a;). 

- _ 7 sin 5 a; 7 

18. dy = — r— aa;. 

a cos 2 a; 

TT /Vl — 2 cos 2 x + cos 4 a;) sin xdx 
Here y = / — 

* i/ cos 2 a; 

= sec x + 2 cos x — -J cos 3 a;. 

19. <fy = cos 7 a-^a;. 

y =. sm x — sin 8 a- + f sin 5 a; -j- | sin 7 x. 



CHAPTER II. 

INTEGRATION OF RATIONAL FRACTIONS. 

136. Rational Fractions. — A fraction whose terms in- 
volve only positive and integral powers of the variable is 
called a rational fraction. Its general form is 

ax m + b x m ~ x + cx m ~ 2 + etc. . . . I . 

dx n + Fx' 1 - 1 + c'x""' r +~Qte'. . . . I" ^ ' 

in which m and n are positive integers, and a, b, . . ., 
a', ~b\ . . . are constants. 

When m is > or = n, (1) may, by common division, be 
reduced to the sum of an integral algebraic expression, and 
a fraction whose denominator will be the same as that of (1) 
and whose numerator will be at least one degree lower than 
the denominator. For example, 

x 5 . 2;r 2 + x 

- x 2 -\- x 



X B _ x 2 + , x + I - ' 38 _ X 2 _j_ x + I 

The former part can be integrated by the method of the 
preceding chapter ; the fractional part may be integrated by 
decomposing it into a series of partial fractions, each of 
which can be integrated separately. There are three cases, 
which will be examined separately. 

137. Case I. — When the denominator can be re- 
solved into n real and unequal factors of the first 
degree. 

f (x\ cJx 
For brevity, let , . denote the rational fraction 

J 0(3) 

whose integral is required, and let (x—a) (x—b) . . . (x— I) 
be the n unequal factors of the denominator. Assume 



DECOMPOSITION OF FRACTIONS. ' 257 

£M = -L- -'+ -*- + -^- . . - 7 - (i) 

^ (#) # — # # — 6 a; — c x — V K ' 

where ^4, B, C, etc., are constants whose values are to be 
determined. 

Clearing (1) effractions, by multiplying each numerator 
by all the denominators except its own, we have 

f(x) = A (x—b)(x— c). . . (x—I) + B (x—a)(x—c) . . . (x—l) 

+ etc. + L (x—a) (x—b) . . . (x—7c), (2) 

which is an identical equation of the {n — l) th degree. To 
find A, B, C, etc., we may perform the operations indicated 
in (2), equate the coefficients of the like powers of x by the 
principle of indeterminate coefficients in Algebra, and solve 
the n resulting equations. The values of A, B, C, etc., thus 
determined, being substituted in (1) and the factor ch intro- 
duced, each term may be easily integrated by known 
methods. 

In practice, however, in this first case, there is a simpler 
method of (in ding the values of A, B, etc., depending upon 
the fact that (2) is true for every value of x. If in (2) we 
make x = a, all the terms in the second, member will re- 
duce to 0, except the first, and we shall have 

f(a) =A(a-b)(a-c)...(a-T), 

A m_ __ = /(«) , 

(a — b) (a — c) ... (a — I) 0' (a) 

In the same way, making x — b, all the terms of (2) 
disappear except the second, giving us 

f(b) = B(b-a)(b-c)...{b-l), 

B = m - m. 

{b — a)(b—c) ...(b-l)~ 0' (J) 



258 CASE L 

Or, in general, the value of L is determined in any one 
of the terms, -, by substituting for a; the corresponding 

root I of <f> (x) in the expression ^rrrk\ i. e., L = ^vUt- 

0' (ar) (Z) 



1. Integrate dy = 



EXAMPLES, 

(a; 2 + 1) ^ 



a*4- 6a; 2 -f- 11a; + 6 

In this example, the roots * of the denominator are found 
by Algebra to be — 1, — 2, — 3. 

.'. a? + 6x 2 + 11a; + 6 = (a; -f 1) (» + 2) (a; + 3). 
Assume 

A +A + A- a) 



a: 3 + 6a; 2 + 11a; -f 6 a; + 1 ' x + 2 a; + 3 
.-. a; 2 + 1 = A (x -f 2) (x + 3) + B (x + 1) (a; + 3) 

+ ff(s + l)(* + 2). 
Making a; = — 1, we have 2 — 2A, .'. A =z 1, 

x = — 2, « « 5 = — B, .'. B = — 5 
«• a; ■ = - 3, " « 10 = 2(7, .-. C = 5. 

Substituting these values of A, B, C, in (1), and nralti 
plying by dx, we have 



y 



O 2 + 1) dx 



X s + 6a- 2 + 11a; -f 6 



J x + 1 «/ a; 4- 2 */ fi 



$e 



+ 1 */ a; + 2 c/a; + 3 
= log (a; + 1) — 5 log (a; + 2) + 5 log (x + 3) 
(a? + 1) (x -f 3) 5 



= los 



(x + 2)* 



* If the factors of the denominator are not easily seen, put it equal to 0, and 
solve the equation for x; the first root may be found by trial, x minus each of the 
several roots in turn will be the factors. (See Algebra.) 



2. Integrate dy 



DECOMPOSITION OF FRACTIONS. 259 

adx 



x 2 — as- 



3. Integrate 



x — a , lx — a 

y = ^^- a = loo »\fx-va' 

\& — 1) dx 

-y - ^ + ^ + i' 

x* _ x + i 



. ■ _ (5a; + 1) $z 

4. Integrate ^ = ^-^j- 

y = log (a — l) 2 (x + 2) 3 . 

5. Integrate dy = ^- 3 - 5, = ^log £±j}. 

138. Case II. — When the denominator can be re- 
solved into n real and equal factors of the first degree. 

fix) 
Let the denominator of the rational fraction —-r con- 

(f>(x) 



C 



tain n factors, 


each equal 


to X 


— a. 


Assume 








/(*) _ 


A 

~ (x — a) n 


1 


B 


cf>(x) ~ 


' (*■ 


- ay- 1 






+ . .. 


L 



(x — a) r 



(x - a) (1) 

Clearing (1) of fractions by multiplying each term by the 
least common multiple of the denominators, we have 

/(*) = A + B {x — a) + C(x - ay 

-f . . . L {x- a y~\ (2) 

which is an identical equation of the {n — l) th degree. To 
find the values of A, B, C, etc., we equate the coefficients 
of the like powers of x, as in the preceding Article, and 
solve the n resulting equations. The values of A, B, C, 



260 CASE II. 

etc., thus determined, being substituted in (1), and the 
factor dx introduced, each term may be easily integrated 
by known methods. 

In this case we cannot find the values of A, B, 0, etc., by the second 
method used in Case I, but have to employ the first. When both 
equal and unequal factors, however, occur in the denominator, both 
methods may be combined to advantage. 



1. Integrate dy 



EXAMPLES. 

(2 — 3^2) dx 



a 2-32* A B c m 

Assume ^T2)-3 = (^T2p + ^+^? + ^r2- (1) 

.-. 2 — 3x 2 = A + Bx + 2B+Ctf + ±0x + 4cC. 

.\ A + 2B + 4:C = 2. (2) 

B + 4(7 = 0. (3) 

G = -3. (4) 

Solving (2), (3), and (4), we get 

A = — 10, B = 12, C = — 3. 

Substituting these values of A, B, and in (1), and 
multiplying by dx, w r e have 

(2 — 3r r 2 ) dx _ lOdx l%dx Sdx 

( X + 2)3 - ( x + 2) 3 + (x + 2p x + 2 



— r ( 2 ~ 3x2) > dx 

(x + 2)3 
5 12 



■/ 



2. Integrate dy 



(x + 2)2 * + 2 
(#2 -f x) dx 



-3 log (a; + 2). 



(a; — 2)2 (x - 1) 



DECOMPOSITION OF FRACTIONS. 261 



Assume 



a; 2 + x _ A B C 

(x — 2f{x — l)~(x — 2)2 + (x — 'I) + x~^~l V-' 

.-. x* + x j= A (x — 1) -i- B(x — 2) (x - 1) 

+ C(x-2y. (2) 

Here we may use the second method of Case I, as follows : 

Making x = 2, we find A = 6. 
x = 1, " « C = 2. 

Substituting in (2) for A and (7 their values, and making 
x = 0, we find 

= _ 6 + 2£ + 8 ; /. B = —1. 
Substituting in (1), and multiplying by dx, we have 



V 



r (x 2 + x) dx 



J (x — 2f (x — 1) 
r 6dx P dx r 2dx 

J ( x _ 2$ — J x _ 2 + J a 7 ^! 



6 



x — 2 



-log (a; -2) + 21og(a;-l). 



6 . (a; — l) 2 




3. Integrate dy = ( J ■ 




Q 

V = - ^73 + 3 log (x - 


-3). 


4-. TnteorafA rf// — —- = dx. 





J x% _ 6^ 2 + 9z * ' 
«/ = log [? (a; - 3) 2 ]*. 



262 case in. 

%. 

139. Case III. — When some of the simple factors 
of the denominator are imaginary. 

The methods given in Arts. 137 and 138 apply to the 
case of imaginary, as well as to real factors ; but as the cor- 
responding partial fractions appear in this case under an 
imaginary form, it is- desirable to give an investigation in 
which the coefficients are all real. Since the denominator 
is real, if it contains imaginary factors, they must enter in 
pairs ; that is, for every factor of the form x ± a + bV— 1, 
there must be another factor of the form x ± a — bV — 1,* 
otherwise the product of the factors would not be real. 
Every pair of conjugate imaginary factors of this form gives 
a real quadratic factor of the form (x ± a) 2 + b\ 

Let the denominator contain n real and equal quadratic 
factors. Assume 

f{x) _ Ax + B Cx + D 

<p (x) " [{x ± af + fl 2 ]" + |> ± of + bY~ l 

+ . . . . KX + L ■ (1) 

^ (x ± a) 2 -j- b* K } 

If we clear (1) of fractions by multiplying each term by 
the least common multiple of the denominators, we shall 
have an identical equation of the (2n — l) th degree. 
Equating the coefficients of the like powers of x, as in the 
two preceding Articles, and solving the 2?i resulting equa- 
tions, we find the values of A, B, C, etc. Substituting 
these values in (1), and introducing the factor dx, we have 
a series of partial fractions, the general form of each being 

(Ax + B) dx 



[(x±ay + W]«> 
in which n is an integer. 

To integrate this expression, put x±a = z ; .*. x = z^fa, 

* Called conjugate imaginary factors. 



EXAMPLES. 263 

dx = dz. (x ± #) 2 = 2 2 . Substituting these values, we 
have, 

r (Ax + -g) ^ / , (^Tifl+^ 

~ J (f -t V) n + «/ (* 2 + ft 2 ) • 

_ ^ r_j^dz_ 

2 (?l - 1) ( 2 2 + J2)*-l + J (# _|_ £2)»> 

(when A = B =F ^4«) ; 

so that the proposed integral is found to depend on the 
integral of this last expression ; and it will be shown in 
Art. 151 that this integral may be made to depend finally 

u P° n /^j^J S iyin g J tan_1 I' ( Art 132 > 3 *) 

EXAM PLES. 

1. Integrate dij — ^~~j- 

The factors of the denominator are 

(x — 1) and (x 2 -f- x + 1). 
therefore assume, 

.-. jc = J2 2 + ^ + J + .gz 2 + j&s — Bx - C. 
.-. A + B = 0; A — J5 + = 1 ; ^4 - C = 0. 

_ r jcdx_ r dx_ r (x — l) ^ 

'"" y " J x> — l~ J z x - 1 «/ ¥ £ 2 + a; + 1 



264= EXAMPLES. 

= iiog(.-i)-i/ < -^>f i; m 

(by changing the form of the denominator.) 

Put x -f -J- = z, then x — 1 = z — §, and $# = dz, and 
the second term of (2) becomes 






<?* 



* 2 + i ■ «/ s 2 + f 3 «/ & + 1 

1 o* 

= - i log (z 2 + I) + — tan-i — (Art. 132, 3.) 

V3 V3 

(by restoring the value of z). 

Substituting in (2), we have, 
V = i [>g (x — 1) — | log (x* + a; + 1) 

+ V3tan-^-±^l. 

V3 J 

2. Integrate dy =; y^p^V 

To find the factors of the denominator, put it = and 
solve with respect to x 2 ; thus, 

#4 + x z _ 2 — 0, or xl + x 2 = 2. 
.-. a; 2 = — I ± f = 1 or — 2. 

... X i + ^2 _ 2 = (3? _ 1) (^ + 2). 

Assume -r— — s s = — — ^ H 7 H rrTT' (1) 

a; 4 + a; 2 — 2 a? + 1 a? — 1 a; 2 -{- 2 v ' 

Hence a; 2 = A (x - 1) (a; 2 + 2) + B (x + 1) (x 2 + 2) 

+ (Ox +- Z>) (3 - 1) (x + 1). (2) 



EXAMPLES. 265 

We may equate the coefficients of the like powers of x, 
to find the values of A, B, G, D, or proceed as follows : 

Making £ =52 — 1, we find A = — i 

x [ = 1, " " B = f 

Substituting these values of ^4 and B in (2) and equating 
the coefficients of x z and x 2 , we have 







6(7=0 and 


6Z> = 4; 






.-. C — and 


7) 2 


••■ 


y = 


6 ,/ a: + 1 + 6 


r dx r dx 

J x -l + 3 J x 2 4-2 




= 


, (x - 1\| a/2 


V2 






EXAMPLES. 


1. 


dy = 


(a; — 1) dx 

x 2 -f 6x + 8 


, (x + 4)* 

V = log ^r« 

9 8 (s + 2)f 


2. 


dy = 


(2a; + 3) <fc 
a; 3 4- a; 2 — 2x 


^ = log 3 / ; Li* 

a; 3 (a; + 2)« 


3. 


dy = 


(Sx 2 -l)dx 
x(x — l)(x + 1) 


?/ = log (x 3 — x). 


4. 


dy = 


adx 

x 2 4- bx 


y-^L x +b Y 


5. 


dy = 


(2 + Sx — 4a: 2 ) f/a; 
4a; — a; 3 








£ = 


log [gi (2 + «)f (2 - a;)]. 


6. 


dy = 


(5x - 2) ^ 
12 





266 EXAMPLES. 

w 7 (a z + brf) dx 
a a l x — x z 

I 7 

«/ = a log $ - — log (« 2 — a; 2 ). 

(3a- _ 5) ^ (3 _ 4)1 

* (x — af (x + a) 



10 . ^ = _£+* + a>* 



a (x — \f (x + 1) 



3 + x . z 2 

+ log 



11. rfv = 



2 (32 _ !) * ^ __ 1)f ^ + 1)f 

x 2 dx 



(x + 2) 2 (x + 1) 
4 

y = ^+~2 + log ( ^ + ^ 






3 , . (x + 3\ s 



^ 



13. dy - ^ _ 2)2 ^ + 3)2 



2/ - - o^r 1 -^ -jh^g(x-2)-- 



2b (x — 2) 125 & v / 25(z + 3) 

+ ih log (a + 3). 

-, . , (z 2 — 4a; + 3)dx 

14. d# = ^ - 2 , ; n • 

* # 3 — 6a; 2 + 92; 

-.. . y = \og[x(x-3?]i 



EXAMPLES. 267 



1 k f l„ _ X ^ X 

» a V - ( X + 2)2 (x + 4)2 



5z + 12 ,__ /a; + 4\* 



i ♦ * e-^y 



XfLX 

!«■ ^ = (a; + 1) (* 2 +1) 



17. % = 



1 . i , i ( x * + i) 1 
y = i tan" 1 a; + log ^- --^- 

dx 



x* _ a; 2 + 2a; — 2 



. (X — 1)3 1 ' . « 

# = log '— — tan 1 -— 

& (^ + 2 )f 3^/2 V2 



18 . ^_ (* + *)& 



^8 _ 2x 2 + x — 2 

(x - 2)t 



19. dy = ^ 



V = 4 tan -1 x + loff 

(a; 2 — a ; + 1) d x 
x 5 + a; 2 -f x -fT 



(a- + 1)1 

?/ = los: - — , — i tan -1 a?. 
J * («J + 1)* 

OA , 9a; 2 + 9a; — 128 7 

20. rfy = -- +- - - dx. 

u ar 3 — 5a; 2 + dx + 9 

(a; _ 3)w 5 

"J = r 1- 

7 I™ i 1 \8 



(x + l) 8 ' a; — 3 
01 ^ 2a^ 

21 - * = { * + 1) (x * +ij- . 

* = log (.—J • 



268 



EXAMPLES. 



22. dy 



(g 8 - 1) dx 
a? — 4 



y = | + log [(* + »)*(* -2)*]. 



23. dy = t^ 



a*fo 



(s+'lHa?+."2)(* + l) 

v = w r (^ + 2)l(^ + l)A- 



4- T 3 o- tan -1 #. 



CHAPTER III. 

INTEGRATION OF IRRATIONAL FUNCTIONS BY 
RATIONALIZATION. 

140. Rationalization. — When an irrational function, 
which does not belong to one of the known elementary 
forms, is to be integrated, we endeavor to rationalize it ; 
that is, to transform it into an equivalent rational function 
of another variable, by suitable substitutions, and integrate 
the resulting functions by known methods. 

141. Function containing only Monomial Surds. — 

When the function contains only monomial surds, it can 
be rationalized by substituting a new variable with an 
exponent equal to the least common multiple of all the 
denominators of the fractional exponents in the given 
function. 

For example, let the expression be of the form, 

m' n' 

ax™ + &E» , 

dy = — -, -,dx. 

a!x c -f- b'x e 



Put x = z mnce ; 

*. x m =■ z m ' nce '. x n — z n ' mce : x c = tfiMw't : x e = z mnce \ 





dx 


= mncez m ™ e -hlz. 






Hence 


dtj : 


(/z r,i'nce 
- a ' z mnc'<i 


4- b$g n,ce 

i y gnnce' 


mncez mnce - 


-Hz\ 


which is 


evidently 


rational. 









270 FUNCTIONS CONTAINING BINOMIAL SURDS. 
1 %\ 

1. Integrate dy = ^dx. (1) 

1 — x* 

Put x = z 6 ; 

then a;* = z 3 , x* = z 2 , and dx = 6z 5 dz ; 

(1 - Z 3 ) 6^ 
- 6 (z G + 2* - Z 3 + 02 _ z + ! _ _J— ) <fe. 

Integrating by known methods, and replacing z by its 
value, we have 

t 7 5 2 1 1 — 1 

3!* X* X* X% X* i ,„ i\i 

7 + 5 - 4 + ^-I+f'- 10 ^ 1 + ^j 

« -r 7 3x*dx 

2. Integrate a?/ = — r 



2#t-_#f 



y=-is 



* 5 - /i * 

%. ± '% d- ^ 'rk ^ ± 16a* - 32 log(2-a*) 



142. Functions containing only Binomial Surds 
of the First Degree. — When the function involves no 

7)1 

surd except one of the form (a + bx) n , it can be rationalized 
as in the last Article, by treating a-\-lx as the variable. 
And therefore can be integrated. ' 

For example, let the expression be of the form, 

dy ' == ~7 = r^ f 
V a + ox 

where n is a positive integer. 

Put a + bx = z 2 ; 

then da; = -r^, x = "7 , and # w == ■- — T^--' 



FUNCTIONS CONTAINING BINOMIAL SURDS. 271 

x n dx 2 (z 2 — a) n dz 



.-. dy = 



Va + bx 



This may be expanded by the Binomial Theorem, and 
each term integrated separately. It is also evident that the 
expression 

x"dx 



(a + lx)* 
can be integrated by the same substitution. 

dx 
1. Integrate dy = 



xv 1 + x 

Put 1 -f x = z 2 ; then t?a; = 2zdz and a; = s? — 1. 
f/a; 2rfc 



.'. rfy = 



<Vl + x & '— 1 



= * _ * 

2 — 1 2 + 1 V ' ' 

2—1 . Vl~+~x — 1 

lUg 

2. Integrate r/^/ 



or y = log 



* + 1 log vr +^ + 1 



(1 + 4a;)* 
Put (1 + 4z) = ^ ; 

A ^ 2C?2 • (2 2 — l) 3 „ , xfi 

then da; = — , a: 3 = v ' - , (1 + 4a;)* = a 5 . 



/. <fy 



' ~ 64 



128 2 4 

f/2. 






272 FUNCTIONS CONTAINING TRINOMIAL SURDS. 

or 



A [<L + _M?_ 3(1+to)i 



» = 128 L 5~ - " (1 +*■"' - ^ITSi! + tQ + uft 

/y>2ll + l f~S rut 

143. Functions of the Form — ~, where n 

is a Positive Integer. \ a + &ac 2 ) 3 

Put a -f W = 2 2 ; 

,, 7 zrtfe 9 2 2 — a , n (z 2 — a) n 

then swfa? = -^-, x 2 = — - — . x 2n = ± — — i_. 

x 2n+1 dx (z 2 — a) n dz 



(a + bx 2 )* 



which may be expanded by the Binomial Theorem, and each 
term integrated separately. It is also evident that the 
expression 

x 2n+l dx 

(a + ~bx 2 y 

can be integrated by the same substitution. 

1. Integrate du = " — • 

6 * Wl - x 2 

Put 1 — x 2 = z 2 ; then xdx = — zdz, x 2 = 1 — z 2 . 

.-. tfy = -^— = - (1 - rf) tfe. 

Vl — ^ 2 

... y . = & " _ z = i (1 - z 2 )i - (1 - x 2 % 

T , , , afffe (2a + Sex 2 ) 

2. Integrate c/?/ = = • «/ = '-z • 

(a + cx 2 )% 3c 2 (a + cx 2 )* 

144. Functions containing only Trinomial Surds 
of the Form Va 4- hx ± oc 2 . 

There are two cases, according as x 2 is + or — . 



FUNCTIONS CONTAINING TRINOMIAL SURDS. 273 

Case I. — When x 2 is +. 

Assume Va + bx + x 2 = z — x ; 

■y* ft 

then a 4- bx = z 2 — 2zx ; .-. x = - — — • 

b + %z 

_ Uf + bz + a) tfs 

and Va + bx + «? = zs r ^= %z + b ■ 

The values of x, dx, and Vci -f- ^ -f x 2 being expressed 
in rational terms of z, the transformed function will be 
rational, and may therefore be integrated and the z replaced 
by its value Vet + bx -f- x 2 — x. 

Case II. — When x 2 is — . 

Let a and (3 be the two roots of the equation 
x 2 — bx — a — ; 
then we have x 2 — bx — a = (x — a) (x — (3); 
/. a + bx — x 2 = — (x — a) (x — (3) 
= (x-a)((3- a). 



Assume 


Vet + bx - 

1 


-X 2 = V(x — a)(f3 — x) 
= (x — a) z; 


.% 


(x — a) (8 - 


-x) = (a? — a) 2 z 2 , 


or 


03- 


- x) = (x — a)z 2 ; 


whence, 




az 2 + )3 
* "" ^ + 1 ' 




.-. 


7 > (« — |3) 2: <r?z 

/^,y2 _i_ ft \ 


and 


V# + bx - 


ICC 4 M 1 

_ (0 — «) z 



z 2 + 1 



274 EXAMPLES. 

The values of x, dx, Va + bx — x 2 , being expressed in 
rational terms of z, the transformed function will be rational. 

1. Integrate dy == 



V a 4- bx + cc 2 



Assume v « + ## + x 2 = 2 — # ; 

then, as in Case I, we have 

a -f bx = z 2 — ~2zx ; .•. x == 

_ 2 (s 2 4- &s + a ) ds 
z 2 4- bz 4- « 



5 4- 2z 



y« 4- &s + # 2 



2z 4- * 

7 — 2 (z 2 + bz 4- «) ^ x (2z 4- 5) 
"'• ** - {]) 4. 2*) 2 x (^ 4- fo 4- a) 

2cZ;Z f?« 



6 4- 2z b , 

2 + * 



y = /r^-= l0 g^ + ') 



2 + * 



== log 
If b = 0, we have 

and if a = 1, we have 

dx 



4- x 4- V« + &e -f- ^ 2 - 



= /—=-=, = \og{x + Va + z 2 ); 
' V « 4- x 2 



= /vm=W(*wi-*). 



EXAMPLES. 2-71 

2dz 
Had we integrated the expression - — ~- without dividing both 

terms of it by 2, we would have found for the integral the following : 
y = log (6 + 22) = log [b + 2x + 2A i /a + bx + x 2 ], which differs from the 
above integral only by the terai, log 2, which is a constant. (See 
Note to Art. 135.) 

dx 
2. Integrate dy = - 

V a + bx — x 2 

Let a and (3 be the roots of x 2 — bx — a = ; then, as in 
Case II. we have 



*Ja -f bx — x 2 = \/{x — a) ((3 — x) = (x — a) z. 

az 2 + 



(13 — x) = (x — a)z 2 ; x = 



z 2 + 1 



(z 2 + l) 2 Z 2 + 1 

_ 2 (« - 3) zdz (z 2 + 1 ) _ 2</z 

""• ^~ («? + 1)2 (0 _ «) « ~ 1+Z 2 ' 

/• ^^ _ r dz 



V a; — a 

tL<r. 

3. Integrate d«/ = 



= — 2 tan" 1 z = — 2 tan" 1 
dx 



x\/\ + .t + £ 2 



Assume vl + fc + a* = z— x, and we have, as in Case I, 

Z 2 — 1 7 2 (Z 2 + 2+1)^2 

dx - 



l + 2z' (l-f2z) 2 

z 2 + z + 1 



Vl + x + £ 2 = 



1 + 2z 



_ 2 (z 2 + z + 1) dz (1 + 2z) (1 + 2z ) _ 2f?z 

■"" rf y = (l-j- 2z) 2 (z 2 + z + 1) (z 2 - 1) ~ z 2 - 1 

= J* * (Art. 137). 

2—1 Z -f 1 V ' 



276 BINOMIAL DIFFERENTIALS. 



■ y - / dX = fJl_ _ /_^_ 

^ Vl + x + a? ^ z - 1 J « + 1 
__ , o , 3 — 1 _ , a? — 1 -4- Vl 4- a; + x 2 

og z + i~ og *"+! + vr+^1^ 

= lo°* . 

° 2 + x -f j^^/i + jp + X 2 

145. Binomial Differentials.— Expressions of the form 
dy = x m (a + lx n Y dx, 

in which m, n, p denote any numbers, positive, negative, or 
fractional, are called binomial differentials. 

This expression can always be reduced to another, in 
which m and n are integers and n positive. 

1st. For if m and n are fractional, and the binomial of the 
form 

1 x-*(a -f- bx^f dx, 

we may substitute for x another variable whose exponent is 
equal to the least common multiple of the denominators of 
the exponents of x; as in Art. 141. We shall then have an 
expression in which the exponents are whole numbers. 
Thus, if we put x = z 6 , we have 

ar$(a + bx^fdx = (Jz~ z (a + M)*> dz, 

in which the exponents of z are whole numbers, and the 
exponent of z within the parenthesis is positive. 

2d. If n be negative, or the binomial of the form 

x m (a + bx- >l ) p dx, 

we may put x = - , and obtain 
z 

x m (a + bx~"Ydx = z~ m \(a + bz n ) p dz, 

in which the exponents of z are whole numbers, and the 
one within the parenthesis is positive. 



CONDITIONS FOR RATIONALIZATION. 277 

3d. If x be in both terms, or the binomial is of the form 

x m (ax 1 -f 1)x h )p dx, 

v, r e may take x l out of the parenthesis, and we shall have 

x m+pt ( a + fa"-*)* dx, 

in which only one of the terms within the parenthesis con- 
tains the variable. 

146. The Conditions under which the General 
Form p 

dy = x m (a 4- hx n ) q dx, 

can be rationalized, any or all of the exponents being frac- 
tional. 

(i.) Assume a + bx 11 = z q . 

p 
Then (a + Ix*)* = z*>. (1) 

Also x = f-^) , 

m 

and tf» = P^-7'- (2) 

• Multiplying (1), (2), and (3) together, we have 

7/1+1 

dy = x™ (a + bz tt )q dx =. S- z p+^ (*^--\ " <fe 3 (4) 

an expression which is rational when ■ — is an integer, 

A n ° 

or 0. 

(£.) Assume « -f lx n = z«#\ 

Then a:" = « (s« — J)-i. (1) 



278 CONDITIONS OF 1NTEGR ABILITY. 

:. x = ah (z<* — b)~K (2) 

,\ x m = a» (z* — 1))~», (3) 

dx = — £ ah {z* — J)-"" 1 &-* dz. (4) 

Multiplying (1) by b, adding a, and taking ^ power, we 
have * 

(a + bx n f = cfl (z* — b) «&. (5) 



Multiplying (3), (4), and (5) together, we have 

a 
n 



p Q ("2+P+l) -(>J1+1+P +1 ) 

x m (a -f bx n )*dx = — ±-a> n q n/ (z' } — b) v " q ' z?+*- l dz, 



9)1 -I- 1 7) 

an expression which is rational when (- - is an in- 

n * no 

teger, or 0.* * 

Therefore there are two cases in which the general bino- 
mial differential can be rationalized : 

1st. When the exponent of the variable without the 
parenthesis increases by unity, is exactly divisible by 
the exponent of the variable ivithin the parenthesis. 

2d. mien the fraction thus formed, increased by 
the exponent of the parenthesis, is an integer. 

Rem. — These two cases are called the conditions of integrabiUty of 
binomial differentials,! and when either of them is fulfilled, the inte- 
gration may be effected. If, in the former case, 1 is a positive 

integer or 0, or in the latter case, + - + 1 is a negative integer 

* The student will observe that Art. 143 is a particular ease of this Article, re- 
sulting from making m an odd positive integer, and n = 2. 

t These are the only cases of the general form which, in the present state of 
analysis, can be made rational. When neither of these conditions is satisfied, the 
expression, if - be a fractional index, is, in general, incapable of integration in a 
finite number of terms. 



EXAMPLES. 279 

or 0, the binomial (g? — a) or (& — b) will have a positive integral ex- 
ponent, and hence can be expanded by the Binomial Theorem, and 

each term integrated separately. But if, in the former case, 1 

Til- 1 - 1 n -■ 

is a negative integer, or in the latter, — h - + 1 is a positive inte- 
ger, the exponent of the binomial (z? — b) will be negative, and the 
form will be reduced to a rational fraction whose denominator is a 
binomial, and hence the integration may be performed by means of 
Chapter II. But as the integration by this method usually gives com- 
plicated results, it is expedient generally not to rationalize in such 
cases, but to integrate by the reduction formulae given in the next 
Chapter. 

1. Integrate dy = x 5 (a -f x 2 )^ dx. 

Here — 1 = 2, a positive integer, and therefore it 

can be integrated by the first method. 

Let . (a + x 2 ) == z. 

Then (a + x 2 )* = z. (1) 

x 6 = (z 3 — # ) 3 , 
a*dz = -I (z 3 — a) 2 z 2 dz. (2) 

Multiplying (1) and (2) together, we have 
dy = x 5 (a -f x 2 ) * dx = f (z 3 — a) 2 zHz. 

3 A o ca o^v 3 /z 10 2az 7 «V\ 

•'• y = 2j {z ^ a + az)dz = 2\i6--r + -r) 

= *(« + *)¥_*!(« + x 2 )i + ^ (a + rf)*. 

2. Integrate rfy = — r = .r~ 4 (l + % % ) * dx, 

x* (1 -J- .r 2 ) 2 

TT w+1 » . — 4 + 1 1 - - .. 

Here ^-+f + 1 = ^--2 + 1= - 1 ' aneg ' 
ative integer, and hence it can be integrated by the second 
method. 



10 EXAMPLES. 






Let (1 + x 2 ) = z 2 x*. ' 






Then x* = (z* - 1)-*. 






x= ( 2 2_l)-i ; 






x~ 4 = (z 2 — l) 2 . 




(1) 


(l + ^) = l + (^^i)-i 






= *(*»— "l)-* 






(1 + z*)-% = z- 1 {? - 1)K 




(2) 


dx = — (2 2 — l)~%zdz. 




(3) 


Multiplying (1), (2), (3) together, we have 






dy = ar 4 (1 4 a 2 )-* dx = — (f — 1) <fe 






•\ y — — J (z 2 — l)dz = z — %z 5 








(2a; 2 - 


-1). 



EXAMPLES. 

1. ^ = MrM*!. (Art. HI.) 

5a;« 

_ , a?* — %x% , 
* 1 + s* 



^£ — 2x — $x% 4 3#t + 2x? — 6x* — 6^ 
+ log (x% + 1)4- tan -1 x%. 



n , 2x? — 3x* 7 
3 - ^ = -r-i r <&• 



3a;3 + a;* 
2/ = 12 (fa?* — fanr 4 J^a;A _ 9^) 
4 1908 [|*ft— Jajlr 4 3gi-- Jyktf* 4 81aj^— 243 log (a^V 4 3)]. 



EXAMPLES. 



281 



A 7 1 +Z 7 

4. ay = -dx. 

1 + x* 

ii 



IT ""8 ~~1 T + 3" 



1 



+ 



2* 



5. % = 



bg / -r^ 1 -) + 2 tan-* (- r ) 



scrfo 



(i- 



(Art. 143.) 



Vi + z + 



VI 



Til 



6. <fy = 



<?» 



aVtf 4- bx 



2 , V« 4- bx — Va 
y = — log 



7. dTjf = 



dx 



(1 + *)• + (1 + *)* 
y = 2 tan" 1 (1 + «)*- 

8. % = 4(a; + Vx 4- 3 4- fyx~+l!)dx. 

y = % {x+SY - 8 (a>+3) + £(3 + 3)* + 3 (*+3)*. 

9. dy = ^t=r- (Art. 143.) 

* Vl + z 2 

jf = i(l + a^i - |(1 4- z 2 ) 1 + (1 + <*)*■ 

T s r fa 3z 2 + 2 



10. <fy = 

11. dy = 



(1 + *)* 

a?dx 
(1 + z 2 ) 1 



y = 



3 (1 4- a^)» 

z 2 + 2 



y == vh * 2 ' 



282 EXAMPLES. 

12. dy = ' '** =. (Art. 144.) 

Vl + x + x 2 

y = log (l + 2z + 2 a/1 + a? + a?). 
(See Art 144, Ex. 1.) 
dx 



13. dy = 



*/ X 2 — x—l 

y '== log (2« — 1 + %^/x 2 — x —I). 



15. d# 



V 2 — a; — x 2 

dx 
*/l 4- a; _ x 2 



16. % = 



y = 2 tan 
dx 



V ^ _ 1 _ iA /5 



V « 2 4- # 2 ^ 2 

(Assume \ / « 4- ^ = z — », etc.) 

y = j log (&c 4- a/« 2 + # 2 # 2 ). 



17. % = 



dx 



xVa 2 4- ^ 2 



1 , /Va 2 + 6 2 z 2 - a\ 

y = i l °8\ — Yx — ) 

1, (a+ Va 2 + ~bW\ 

r -« io n — x — ) 



18. dy 



_ (2x + x 2 )? dx 



x 4 



= log[. 



# 4- J 4- V 2a; 4- a; 2 



x + V2a: 4- re 2 - 



19. dy = 



EXAMPLES. 283 

dx 



(1 + a; 2 ) Vl — x 2 
y __ -— tan -1 I — ___= J 



adx 
20. dy = 



V2ax + a; 2 
y = a log (a + « + a/2«z + a; 2 ). 

c?^ _____ 

2L f ^ = V4^^' ?=.!<*(** + <•_»-?). 
(Compare with Ex. 16.) 
2dx 

(Compare with Ex. 20.) 

23. dy = afi(2 + 3a?)*. (Art. 146.) 

y = h\r ± f^ - *< 2 + s *>* + *( w)t) 

24. <&/ = a 3 (a + fo; 2 )^ efo. 

, 7 0N 5 /5to 2 — 2a\ 

25. dy = a? (a + fa*)* <fo. 

y = i ^(3^-2«)(a + fa*)t. 

26. c?y = X s (a — &)-? dx. 

V = - 1 (a - a*)* (2« + z 2 )- 

27. dy = *L_ . „_ L_/l 2 \ 

a* (1 + j*)t y - ~ (1 + a^Uz + ®T 

28. dy = a(l + x*)~% dx. y = — — -. 

(1 + z 2 )* 



284 EXAMPLES. 



29. dy = x~* (1 — 2x 2 )~* dx. 



— *&*-+ 



30. J?/ = (1 + x^x^dx. 



_ (3a? — 2) (1 + 3 *)* 
^ "" 15 



5 



31. tfj/ = x~ 2 (a + a 3 )-3 & 

(3^ 3 + 2d) 
y = ~ 2d>x (a + x*)%' 

32. dy = X s (a 2 + xrf dx. 

y = .^ (a 2 + z 2 )t (4z 2 — 3a 2 ). 

33. dy = x 5 (a + foc 2 )^ *fo. 

3z 5 / * 6 • ^ 3 , a?\ 

in which z = (« -f 5a; 2 ) £. 

34. rfy = < a + fe )* <fe . 



+ di log .-Z_- — • 

va + J» + v« 

35. dy = (a 2 + z 2 )* cfo. 



= -V+^L* + « 2 log[a; + {fla . + a , ¥] . 



»- ■ a '2 



CHAPTER IV. 

INTEGRATION BY SUCCESSIVE REDUCTIONS. 

147. Formulae of Reduction. — When a binomial dif- 
ferential satisfies either of the conditions of integrability, it 
can be rationalized and integrated, as in the last chapter. 
But, instead of rationalizing the integral directly, it may 
be reduced to others of a simpler kind, and finally be made 
to depend upon forms whose integrals are fundamental, or 
have already been determined. This method is called 
integration by successive reduction, and is the process 
which in practice is generally the most convenient. It is 
effected by formulae of reduction. These formulae are 
obtained by applying another, known as the formula for 
integration hy parts, and which is deduced directly as 
follows : 

Since d (uv) = udv -f- vdu, (Art. 16) 

we have uv = I udv + / vdu ; 

/ udv = uv — / vdu ; 

a formula in which the integral of udv depends upon 
that of vdu. 

148. To find a formula for diminishing the expo- 
nent of x without the parenthesis by the exponent 
of x within, in the general binomial form 



fx m (a + bx n )p dx. 



286 SPECIAL FORMULA OF REDUCTION, 

Let y = I x m (a + bx 11 )* dx = I udv 

= uv*h vdu ; (1) 

and put dv = x n ~ l (a + bx n )v dx and u == x m ~ n+1 . 

Then v = —-, — -^rv 

nb (p + 1) 

and du = (m — w + 1) x m ~ n dx. 

Substituting these values of u, v, du, dv, in (1), we have 

f* , 7 x 7 x m - n+l (a + bx n )p +1 
y = J * (a + far); efc ;= ^ + ^ 

«& (_P + 1) «/ V ' W 

This formula diminishes the exponent m by w as was 
desired, but it increased the exponent p by 1, which is 
generally an objection. We must therefore change the 
last term in (2) into an expression in which p shall not be 
increased. 

Now x m -" (a + bx n Y +1 = x m ~ n (a + bx")? (a + bx n ) 

= ax m ~ n (a 4- bx n )p + bx m (a + baf i y J 

which in (2) gives 

y = J *r (a + bey dx = --S—-J- 

n (p + 1) «/ v y 

Transposing the last term to the first member and redu- 
cing, we have 



SPECIAL FORMULA OF REDUCTION. 287 

( np + m + 1) f , wy dx _ *"-"-> + W n 



m — n + 



nb (p + 1) 
Therefore we have 



- a Jx m ~ n (a + ox 1i y dx. 



y z= J x m (a + bx n Ydx 
x m - n+l (a + bx n Y^ — {m— n.+ l)a j x m - n (a + bx n ) p dx 



b (np -f m + 1) 
tvhich is the formula required. 



\<A) 



149. To find a formula for increasing the exponent 
of x without the parenthesis by the exponent of x 
within, in the general binomial form 



y — fx- m (a + bx n )*> 



dx. 



Clearing (A) of fractions, transposing the first member 
to the second, and the last term of the second to the first, 
and dividing by (m — n -f- 1) a, we have 

fx m ~ n (a -f- bx n )p dx 



a (m — n-\-l) 



a) 



Writing — m for m — n, and therefore — m + n for m, 
(1) becomes 

y = / x~ m (a + bx n Y dx 

x- n +\a + bx n Y n + h(m - np -n-1) / V- n,+n (tf + bx 1i ydx 
= -a(m-l)~ ~ ;(jB) 

which is the formula required. 



288 SPECIAL FORMULA OF REDUCTION. 

150. To find a formula for diminishing the expo- 
nent of the parenthesis by 1, in the general bino- 
mial form 

y = J x in (a -)- bx n )p dx. 

j x m (a + bx 1l )p dx — j x m (a -4 bx n ) p ~ l {a + bx 11 ) dx 

— a J x m (a + bx n Y~ l dx + b J x m+n (a + bx n y~ l dx. (1) 

By formula (A), we have for the last term of (1), by writ- 
ing m + n for m and p — 1 for j?, 

fx" l+n (a + bx n Y~ l dx 

x m+1 (a -f bx n Y — (m + 1) a f * x m (a + bx n y~ l dx 
b [n (p — 1) + m -f n + 1] 
which in (1) gives 

y — j x m (a + 5^ M )^ dx = a J x m (a + bx n y~ l dx 

x m+i ( a + ^p _ ( m 4. i) ^ y a;»i ( a _j_ fa*)*-* dx 
(np -+- m 4- 1) 

Therefore, uniting the first and third terms of the second 
member, we have 

y — J x m (a -f bx n y dx 
x m+l (a + bx n y + anp f x m (a + bx n Y~ l dx 



np + m + 1 
ivhich is the formula required. 



(C) 



SPECIAL FORMULJE OF REDUCTION. 289 

151. To find a formula for increasing the exponent 
of the parenthesis by 1, in the general binomial form 

y = j x m (a + bx n )~p dx. 

By transposing and reducing (C), as we did (A) to find 
(B), we have 

fx m (a + bx")p~ l dx 

x m+l (a 4- hx n Y — (112) + m + !) I xm ( a + bx n )p dx 

= J - (1) 

— a up 

Writing — p for p — 1, and therefore — ^ + 1 for ^;, 
(1) becomes 

ij — J x" x (a + fee") - * da; 

a^+^a + ^r')-p* 1 — (m + n -f 1 — mjp) / x m (a + hx n )~p +l dx 

= ^uT^jf ' W) 

which is the required formula. 

Remark. — A careful examination of the process of reduction by 
these formulae, will give a clearer insight into the method than can be 
given by any general rules. We therefore proceed at once to exam- 
ples for illustration, and shall then leave it to the industry and inge- 
nuity of the student to apply the method to the different cases that 
he may meet with. 

EXAMPLES. 

x 1,l dx 



1. Intesrate dy = 

Here y = / x m (a 2 — x 2 )~^ dx, 

a form which corresponds to 

f x m (a -f- bx n Y dz, 
13 



290 EXAMPLES. 

We see that by applying formula (A) we may diminish 
m by 2, and by continued applications of this formula, we 
can reduce m to or 1 according as it is even or odd, so 
that the integral will finally depend upon 

/dx . x 

— r = sin -1 - , when m is even ; 

Va 2 -x 2 a 

/xdx i 

-— — = — (a 2 — x 2 )?, when m is odd. 

Va 2 -x 2 

Making m = m, a = a 2 , b = — 1, n = 2, p = — \, 
we have from formula (A), 

y = J x m (a 2 — x 2 )~i dx 

x m-2+i ( a 2 _ x 2^ _ a 2 ( m — 2 -f- 1) Cx m - 2 {a 2 — x 2 )~? dx 

= - P (- \) + m + 1] 

x m ~ l (a 2 — x 2 )? 
m 

(m — 1) a 2 fx m ~ 2 (a 2 — x 2 )"* dx 

+ fL (i) 

When m = 2, (1) becomes 

f xhlx x i , a 2 . a; 

When m = 3, (1) becomes 

= C_ jM±_ = _ j^ ^ _ ^ _ j a2 (fl a _ ^| 

^ y a 2 — # 2 
= _ i ( fl 2 _ gp)i (^ + 2« 2 ). 

When m = 6, (1) becomes, by applying (.4) twice in 
succession, 



EXAMPLES. 291 

xhlx 



= /: 



Va 2 — x l 

(which the student may show.) 

_ T , 7 a^a; 

2. Integrate ay = — = 



V^ 3 + % 2 
Here . y = J x m (a 2 + x 2 )~? dx. 

Making m = m, a = a 2 , b = 1, w = 2, /? = — \, 
we have from (^4), 

y = / a m (« 2 + a?) "* <fo 

= a ^ — / x" 1 2 (a 2 + a; 2 ) * ##. (1) 

By continued applications of this formula, the integral 
will finally depend on 

/ — - = log (x -f Va 2 4- a; 2 ), when w is even, 

«- Va 2 + z 2 

/xclx 1 

— — = (a 2 + x 2 )* 9 when wi is odd. 
yet 2 + £ 2 

3. Integrate £??/ == 



« m (a 2 — x 2 )? 

Here V — I x ~ m { a% — ^ 2 )"~^ *&, 

from which we* see that by applying (B) we may increase m 
by 2, and by continued applications of (B), we may reduce 
m to or 1, according as it is even or odd, making the 
integral finally depend on a known form. 



292 APPLICATIONS OF FORMULAS. 

Making m = m, a = a 2 , b = — 1, n .= 2, p = — |, 

(B) gives us 

y = J %- m (a 2 — x 2 )~i dx 

x -m+i ( fl 2 _ ^i _ ( m + i __ 2 — 1) fx- m+2 (a 2 —x 2 y? dx 

— a 2 (m — 1) 
_ (a 2 — x 2 )i (m — 2) r dx 

" ( m _ 1) ^—1 + ( m _ 1) fl 2 «/ ^ m _ 2 (fl j _ ^ 2) 1* V ) 

When m = 2, (1) becomes 

r dx _ _ (a 2 — x 2 )i . 

V " J x 2 Va^^x 2 ~ &* 

(since the last term disappears.) 
When m = 3, (1) becomes 

dx (a* — x 2 )i 1 f dx 



y 



r dx _ _ (a 2 — x 2 p j^ r 



sB»Va*— a? 2a ^ 2 2 « 2t/ a?Va 8 — a* 

_ _ a/« 2 — £ 2 1 -, « — Vet 2 — x 2 
2fl¥~" + 2a 3 ° g 5 

(Ex. 1 7 of Art. 146.) 
4. Integrate dy = (a 2 — x?)'» dx, when n is odd. 
Here we see that by applying (C) we may diminish 

- by 1, and by continued applications of (C) we can reduce 

- to — J, making the integral depend finally upon a 
known form. 

Vh 

Making m = 0, a = a 2 , b = — 1, w = 2, ^? = 5 , (C) 
giyes us 

2/ = ) {a 2 — x 2 )? dx 

x {a 2 — x 2 f~ + na 2 f\a 2 — x 2 )^ 1 dx 
= _ J — (1) 



APPLICATIONS OF FORMULA, 293 

When n = 1, (1) becomes 

y = J ( a - x )* dx = 2 + 2 sin a* 

dx 

5. Integrate d# = — , when n is odd. 

(a 2 — x 2 )j 

Here V = J ^ ~" ^J -15 ^» 

from which we see that by applying (D) we may increase 

ft 
the exponent - by 1, and by continued applications of (D) 

we can reduce : - to — j£ making the integral depend 

finally on a known form. 

Makii 
gives us 



ft 

Making m = 0, a = a 2 , b = — 1, n = 2, p = -, (Z>) 



y =z I [a 2 — a; 2 ) 2 d?# 



2a 2 



G-») 



(» - 2) a 2 (« 2 - X 2 ^ ( n - 2) « 2 «/ ( a 2 _ ^-1 V ' 

When n = 3, (1) becomes 



-/ 



dx 



(a 2 - x 2 )i a 2 (a 2 - x 2 )% 



6. Integrate <#?/ = — -= 

v2ax — 



Here y z= Jx m (2ax — a£)-£ dfo = f x m -^a—x)-^dx y 
which may be reduced by (.4) to a known form. 



294 APPLICATIONS OF FORMULAE. 

Making m = m — %, a = 2a, b = — 1, n = 1, p = —\, 
(A) gives us 

/x m dx 
— — 
V2ax — £ 2 

#"*-£ (2a — x)i — 2a (w — £) AfHI (2a — x)~~l dx 

— m 

x m ~ l y x 5 , (2m — 1) a r x m - l dx ._ 

= V%ax — x 2 + ± — I — (1) 

™ w» ^ V2ax-x 2 

When m = 2, (1) becomes 

/aftfcc x -\- da rz - 9 
— = -= — V 2az — x 2 . 
a/o„~ ^2 2 



V%ax — x 2 % 

4 



K*/ 



V 2a# — a; 2 



= ^ — v2a# — r 3 + fa 2 vers * — 

2 2 a 

xfidx 
7. Integrate a'y = 



VI — a 2 

1.3-5 . 



/z 5 1 • 5 , 1 • 3 • 5 \ /- 5 1.3. 



-1. 



sm *#. 



8. Integrate dy = 



a 4 Va 4- £a; 2 

* = (- m + Si) V * TJ * K 

9. Integrate dy = (1 — #*)$ a#. 
y = £3 (1 — z 2 )^ 4 P (1 — £ 2 )* 4- 1 sin" 1 #. 



10. Integrate dy = 



(1 + x 2 f 



+ s • t\ 5\ + 5 tan x a;. 



4(1 -fz 2 ) 2 ' 8 (1 4-a 2 ) ' S 



11. Integrate dy — 



LOGARITHMIC FUNCTIONS. 295 

xhlx 



<S/%ax — x 2 
g- -f- f £ fl -f | ■ fa 2 ] V2ra — a: 2 + 4 • |« 3 vers" 1 -. 

12. Integrate dy = =• 

6 * & yi _ X 2 

/ 1 1 • 3 \ 7l 2 1 • 3 . 1 + a/I"^" 2 

These integrals might be determined by one or other of 
the methods of Chapter III ? but the process of integration 
by reduction leads to a result more convenient and better 
suited in most cases for finding the definite integrals.* 

LOGARITHMIC FUNCTIONS. 

152. Reduction of the Form / X (log x) n dx, in 
which X is an Algebraic Function of x. 

Put Xdx = dv and log" x = n. 

/dx 
Xdx and du = n log" -1 x —— 

Substituting in / ndv = uv — I vdu, (Art, 147) 

we have y = I X log n xdx 

= log" x J Xdx — fn log' 1 " 1 x J' (Xdx) — ; 

or by making / (Xdx) = X l , 

we have y = J X log n xdx 

— l log" -1 2$s ; 

* For a discussion of definite integrals, see Chap. V. 



296 EXAMPLES. 

which diminishes the exponent of log x by 1, wherever it 
is possible to integrate the form / Xclx. By continued 

applications of this formula, when n is a positive integer, 
we can reduce n to so that the integral will finally 
depend on 

^ x 

Sch. — A useful case of this general form is that in which 
X — x m , the form then being 



y = I x m log 71 xdx ; 



and the formula of reduction becomes 
x m log n xdx == log" x 



; — - I x m log"" 1 xdx. 



m '+ 

by means of which the final integral, when n is a positive 
integer, becomes. 

r _ x rn+1 

/ x m dx — -• 

t/ m + 1 

EXAMPLES. 

1. Integrate dy = x 4 log 2 xdx. 

Making m = 4, and n = 2, we have 

y — I x* log 2 sw/a; 

= ^?-|/^logA (1) 

Making w = 4 and w = 1, we have 

y* iog^ & = slip - \j-tHA (-f ), 



LOGARITHMIC FUNCTIONS. 297 

which substituted in (1) gives us 

/* 410 7 x 5 log 2 x 2 /x 5 log x 1 2 5 \ 

•?/ z= / X i log 2 £ $C = z 5 ^ ~ • - ) 

J J ° o o \ 5 55/ 

a 3 

- x i. , - , a? log # t/# 

2. Integrate ay = — 

Put — = dv and log x = u : 

Va 2 + x 2 

/JfY' 

then v = v« 2 4- £ 2 and ^ = — 

0! 



/xhsxdx , _ ox i . / > v« 2 + £ 2 7 

— 1==- = (a» + x?)* logx-J v --t- dx 
Va 2 + x 2 " x 

= (« 2 + a^t \ogx- J —==- J -—= 

V X^ a 2 _|_ ^2 «/ y a 2 _j_ ^2 

= (« 2 + ^)i fogs + a log ( ^ + ^f + ) - V^T^. 

(See Ex. 17, Art. 146.) 
log x dx 



3. Integrate dy 



(1 + *) 2 

y = y^x log * ~~ log ^ + * 



153. Reduction 



log" x 



r r.r™ fix 
of JVo 

Put x m+1 — u, ,—- — — = dv: 



1_ dx 
log" # a; 



then ^w = (m + 1) a m ^x 

and v = 



(71 — 1) log 71-1 X 



298 • LOGARITHMIC FUNCTIONS. 



. _ {*#* dx _ x ,n+1 -«z + l r x m dx 

y ~~ J \og n x~ (n—l)log n ~ l x (n—l)J log" -1 a;' 

by means of which the final integral, when n is a positive 
integer, becomes 

Px m dx 

J log x 9 

beyond which the reduction cannot be carried, for when 
n = l the formula ceases to apply. We may, however, ex- 
press this final integral in a simpler form ; thus, 

Put z = x m+} ; 

then dz = (m -f- 1) x m dx and log z = (m + 1) log x. 

Cx m dx _ f dz 
J log x ~ J log z ' 

an expression which, simple as it appears, has never yet 
been integrated, except by series, which gives only an 
approximate result. 

Ex. 1. Integrate dy = ; — =~ • 
& J log 2 x 

Here m = 4 and n = 2; therefore the formula gives us 

/x i dx x 5 rtfdx 

log 2 x ~ log x J log X 

Put z = x 5 ; then dz = 5x*dx and log z = 5 log x; 

,. „ Px*dx P dz 

therefore / . = / , 

J log x J log z 

Now put log z = u ; then z = e H and dz == e"df?^. 

/dz P e"du 

log 2 ~~ J u 

»/(l + .+.£ + £ + «*}£<**«) 



EXPONENTIAL FUNCTIONS. 299 



= log u + u + ^ + ^ + etc. 

= log (log a?) + logo: 5 + ^log 2 ^ 5 + -^^- + etc. 



eta 
a; 



" ^ ~ t/ log 2 

+ 5 



log a; 



log (log a?) + log a* + -Sjp- 



log 3 ^ 5 

+ *f + etc J- 



(See Strong's Calculus, p. 392 ; also, Young's Int. Cal., 
pp. 52 and 53.) 

EXPONENTIAL FUNCTIONS. 

154. Reduction of the Form / a™* oc n doc. 

Put a mt dx = dv and x n = u ; 

a"" 

then v = — . and du = nx n ~ l dx. 

m log a 



y = I a mx x n dx = — ; = / a" 

^ t/ m log « w loff « «/ 



dr. 



By successive applications of this formula, when n is a 
positive integer, it can be finally reduced to 0, and the in- 

/a mr 
a nix dx = — : 
m log a ■ 

Only a very few of the logarithmic and exponential func- 
tions can be integrated by any general method at present 
known, except by the method of series, which furnishes 
only an approximation, and should therefore be resorted to 
only when exact methods fail. 



300 EXPONENTIAL FUNCTIONS. 

EXAMPLES. 

1. Integrate cly = a x x?dx. 

Here m = 1 and n == 3 ; therefore, from the above for- 
mula, we have 

y = I a x xhlx 

— / a x x 2 dx; (by repeating the process) 



a x x° 



log a log 
fl^ic 3 3 / a x x 2 



J a x xdx\ ; 



log a log « \log a log a 

(by repeating the process) 
a x x* 3a*x* 6 / a x x 1 \ 

J I flX I 

log « log 2 « log 2 « \log a log 2 a / 

a* / _ _3a?_ 0>x 6 X 

~ log a V log « log 2 # log 3 «/ 

2. Integrate dy = x B e ax dx. 

/rt 35 (Zoo 
— — , when mis a posi- 
tive Integer. 

Put x~ m dx = dv, a x = u; 

then v = and du — a x log a dx. 

m — 1 

/*a x dx _ a x log « r cFdx 

V — J x »i ~ j^~_ i) x m ~ l m — \J x"'~ L ' 

by means of which the final integral becomes 

fcFdx 
J x 



TRIG 0N02IETRIC FUNCTIONS. 



301 



which does not admit of integration in finite terms, but 
may be expressed in a series, and each term integrated sepa- 
rately. (See Lacroix, Calcnl Integra', Vol. II, p. 91.) 

-n 1 T 4- 4 7 

Ex. 1. Integrate ay = — j- 

. X 

By the formula just found, we have 



Pdx 



rc x d. 

= --+/ 

X J 



f* 



■¥ 



ri x di 



/v2 nr& 

1 + *+a+0 + eto - 



% (Art. 62) 



6"^ X X° 

- - ~ + log x + x + a + sna + etc. 



TRIGONOMETRIC FUNCTIONS. 

156. Cases in which sin" 1 cos H is immediately 
Integrable. — The value of this integral can be found im- 
mediately when either in or n, or both, are odd positive 
integers ; and also when in + il is an even negative 
integer. 

1st. Let m = 2r + 1 ; then 

^ sin" 1 cos" Odd = j '(sin 0) 2 ' +1 cos" dB 

= f (1 — cos 2 ey cos" sin dd 

== — y (i — cos 2 ey cos" d> cos 0, 

an expression in which the binomial (1 — cos 2 0) r can be 
expanded, and each term integrated immediately. In like 
manner, if the exponent of cos be an odd integer, we may 
assume n = 2r -f- 1, etc. 



803 EXAMPLES. 

2d. Let m -f- n = — %r ; then 
J*mi m e cos" (9 a 7 = ftsm m Q (cos0)" +wl ^0 
= yW^ (sec0) 2r d0 

= ytan" 1 (1 + tan 2 0)'- 1 tf • tan 0, 

each term of which, after expansion, can be immediately 
integrated. 

e x amples. 

1. dy = sin 2 cos 3 Odd. 

Here y = y*sin 2 cos 3 d0 

= J sin 2 0(1 — sin 2 0) d- sin 
= i sin 3 0—| sin 5 0. 

O 7 Sm2<9 7A 

2. dy = — -7-z dd. 



cos 4 

sin 2 



Here y = /— — af0 = Aan 2 sec 2 
^ «/ cos 4 t/ 



f/0 



= J tan 3 0. 



3. dy = sin 3 cos 4 d dd. y = —\ cos 5 + \ cos 7 0. 

4. fZy — sm5 ^ cos5 ^ dd. 

y — — -jV cos 6 (sin 4 — J sin 2 + |). 

5. dy = sin 3 cos 7 cZ0. y = ■£> cosl ° # — t cos8 & 

m'n 2 
6 - ^ = ^Sfl^- 2/-itan 3 + itan 5 0o 



7. dy = 



cos 6 

£70 



sin cos 5 



FORMULAE OF REDUCTION. 303 

_ P sects d0 m 4- tap 2 fl) 2 sec 2 6 dd 

Here y _ J ^q—^ -J tan e 

= log (tan 6) + tan 2 + J tan 4 0. 
8. dy = . a , ft ' ■ 

sma cos a a 
Let a: = tan 0; 

1 # 

then cos = — T > sin = 



and * = ^===, 

... f = /-.^-r = 7^+f* 

^ sin* cos* «/ a?f 

= £zt_- =. f tan*0 



^ tan*0 

9 cos 3 ^4 

in 

10. d# = — — # = tan (9 + f tan 3 + £ tan 5 0. 



157. Formulae of Reduction for 

sin m cos" dd. 



/• 



When neither of the above mentioned conditions as to m 
and n is fulfilled, the integration of this expression can be 
obtained only by aid of successive reduction. 

We might produce formulas for reducing the expression 
sin" 1 cos" directly ;* but, as it would carry us beyond 
the limits of this book, we prefer to effect the integration 
by transforming the given expression into an equivalent 
algebraic form, and then reducing by one or more of the 

* See Price, Lacroix, Williamson, Todhmiter, Courtenay, etc. 



304 EXAMPLES. 

formulae (A), (B), (C), (B). Thus, put siu = x, then 
sin"' = x m , cos = (1 — a 3 )*, cos" = (1 — z 2 )l, and 
<70 = (1 t- a*)~* g& 

.-. ^ — /sin"' cos" 0rt0 = y m" 1 (1 — a?)~*~dx. 

or we may put cos = a? 3 and get 

sin™ cos" Odd = J — x n (1 — a^) 1 dx; 

either of which may be reduced by the above formulae. 

This process will always effect the integration when m 
and n are either positive or negative integers, and often 
when they are fractions. The method is exhibited by the 
following examples. 





EXAMPLES. 


1. 


dy = sin 6 Odd. 


Put 


sin = x f 


then 


dd = (1 - x*fi dx. 



.-. y =z y s in G Odd = J*x 6 (1 — a£)~i dx 
/x 5 5x 3 3-ox\ ^ 9 .i , 3-5 . 



sm * X 



(by Ex. 7, Art. 151); 

cos0 , . , . _ 5 . , . 5-3 . m , 5-3 
= --- (sm^ + ^sm3 + --sm0) + £-_ 0. 

7 f ^ 

2. dy = 



sm 5 

Put sin = #, 

then dd = (1 — a; 2 )"* tfa?. 



INTEGRATION. 305 



«/ sin 5 J x ' 



-( 



' >-»>vr^f-t?i«, 1 + - 



(by Ex. 12, Art. 151); 

cos 



4 



/I 3 \ 1-3, ± ,„ 



/ • , 1 + cos sin . . _ x 

(since — log . a = log — = log tan W.) 

v ° sin ° 1 + cos & ^ y 



3. dy = sin 4 

s 4 
4 



y = ^ (sin3 + | sin 0) + f 0. 



(See Ex. 10, Art. 135.) 

4. % =s cos 4 0t/0. 

sin cos 3 , . rt 
?/ = — f- | sm 6 cos + {(9. 

(See Ex. 9, Art. 135.) 

158. Integration of sin m cos n dd in terms of 
the sines and cosines of the multiple arcs, when m 
and n are positive integers. 

The above integrations have been effected in terms of the 
powers of the trigonometric functions. When m and n 
are positive integers, the integration may be effected with- 
out introducing any powers of the trigonometric functions 
by converting the powers of sines, cosines, etc., into the 
sines and cosines of multiple arcs, before the integration is 
performed. The numerical results obtained by this pro- 
cess are more easily calculated than from the powers. 

Three transformations can always be made by the use of 
the three trigonometric formulae. 



306 EXAMPLES. 

(1.) sin a sin b = \ cos (a — b) — \ cos (a + b). 
(2.) sin a cos & = J sin (a + J) + j. s i n ( a _ &). 
(5.) cos a cos b = \ cos (a + b) + £ cos (a — b). 

EXAMPLES. 

1. cfy = sin 3 cos 2 0^0. 

Here sin 3 cos 2 = sin (sin cos 0) 2 

= sin (J sin 20) 2 [by (2)] 

= I sin (sin 2 20) 

. . Q (1 — cos 40\ __ 

= i sin ( j [by (1)] 

= -J- sin — i sin cos 40 

= | sin — i (i sin 50 — J sin 30) 

[by (2)] 
= -| sin — ^ sin 50 + ^ sin 30 

,\ y = /'sin 3 cos 2 Odd 

= f{\ sin Odd — ^ sin 50^0 + -^ sin 30^0) 

= — -| cos -|- -fa cos 50 — ^ cos 30. 

2. ofy = sin 3 cos 3 0^0. 

2/ = — A c °s 20 -f yfo- cos 60. 

3. dy = sin 3 0J0. 

?/ == T ^ cos 30 + f cos 0. 

4. dy = cos 3 Odd. 

y = ^ sin 30 + £ sin 0. 



FORMULJE OF REDUCTION. 307 

159. Reduction of the Form 

/ x n cos ax dx. 



Put 


u = x v , 


and 


dv = cos ax dx ; 


then 


du = nx 1l ~ l dx, 


and 


1 . 

v = - sm az. 


.\ y = J % n cos ax 


= - x n sin az / x n ~ l sin a# dfo. 

a a*J 


Again, put 


u = af -1 , 


and 


dv == sin # # eta ; 


then 


6?w = (» — 1) x n ~ 2 dx, 


and 


1 

v = cos ax. 

a 



/x n ~ x cos ax dx = x n ~ l cos ax 
a 

n — 1 C _* 

H / a* -2 cos tf# «#. 

.\ y =■ ) x n cos axdx =. - x n sin «# 

( # n_1 cos ## H / # n ~ 2 cos ax dx) 

a \ a at/ J 



. 2 cos axdx. 



x n ~ l {ax sin ax -f- n cos ax) n (n — 1) 
a 

The formula of reduction for x n sin ax can be obtained 
in like manner. 

EXAM PLE, 

1. dy = x % cos x dx. 

y = x 5 sin x -f 3# 2 cos a — 62 sin x — 6 cos x. 



308 FORMULA OF REDUCTION. 

160. Reduction of the Form 

/ e ax cos" x dx. 



Put 


U = COS i( X, 


and 


dv — e ai dx; 


then 


du = — 11 cos'" -1 x sin x dx, 


and 


e ax 
v ■ = — • 


.-. y — 1 e ax cos M 


, . e a * cos n a; 


J/ U/Jy — 

a 




_j / e ax COg n-l # gj n # ^ (]_) 


Again, put 


« = cos' 1-1 a? sin a**, 


and 


6?v = e ax ; 


then 


^?'^ == — (n — 1) cos n-2 a; sin 2 x dx 




+ cos" a: dx, 


and 


e ax 

v = — • 



/ e a * cos" -1 a; sin #ffc 

= - e ax cos 71-1 a; sin a; — - / e ax \— (n — 1) cos w ~ 2 a? sin 2 x 

+ cos" x] dx 

== - e ax cos' 1-1 a; sin a? + / e ax cos n ~*xdx 

a a v 

I e ax cos 2 x dx. (Since sin 2 x = 1 — cos 2 x.) 

Substituting in (1), and transposing and solving for 
/ e ax cos" x dx, we get 

/.. e ax cos" -1 x (a cos x + n sin a;) 
<f x cos" a? da: as - T - — 5 ' 
a 2 -f « 2 

' J V cos"" 2 a; <fo ; (2) 



a* -j- ^ 



EXAMPLES. 309 

which diminishes the exponent of cos x by 2. By con- 
tinued applications of this formula, we can reduce n to 
or 1, so that the integral will depend finally on 



/ 



e , 
e ax ax = — , when n is even ; 
a 



or / e ax sin xdx 9 when n is odd. 

(2) gives the value of / e ax sin x ax without an integra- 
tion, since the. last term then contains the factor n — 1 
= 1 — 1 = 0, and therefore that term disappears. 

The reduction of / e" sin x dx can be obtained in like 
manner. 

EXAM PLES. 

1. dy = e ax cos x dx. 

e ° X t , ■ \ 

y — &~+i ^ cos x + sm ^' 

2. dy = e"* cos 2 x dx. 

_ e ax cos x(a cos x -f 2 sin #) 2 e* 1 * 

y ~~ 4 -fa 2 + 4+~a* ' "5" 

161. Integration of the Forms 

/(a?) sin -1 x $a?, /(ac) tan -1 x dx, etc. 

Integrals of these forms must be determined by the 
formula for integration by parts (Art. 147) ; the method 
is best explained by examples. 

EXAM PLES. 

1. dy = sin -1 x dx. 

Put dv = dx, and u = sin -1 x; 

then v = x. and du = — — 

Vl - & 



310 EXAMPLES. 

.*. y = I &\rr l xdx 

. , f xdx 

= x sin -1 x — / — 

= a; sin -1 x -f (1 — a= 2 )i 



a: 2 tan -1 acta 
3. * = ____. 

y-w . 7 JU iXJL ~ Lit*/ 

Put £W = - — - r> = dX — 



1 + X 1 1 + X 1 ' 

and tk = tan -1 x ; 

then v = x — tan -1 #, 

and aw = 5 - 

1 -\- x 2 

. ,, t .«, Pi xdx tan -1 £^oA 
.-. y = x tan -1 a; — (tan -1 an 2 — / ( - n = ■=- ) 

* v ■ «/ \l + ^ 1 + x 2 I 

= x tan" 1 x — (tan" 1 a:) 2 — % log (1 + a; 2 ) + \ (tan" 1 z) a 
= x tan" 1 a? — £ (tan" 1 x) 2 — | log (1 + x 2 ). 

3. dy = x 2 sin -1 a: dx. 

y = - sin" 1 x — -J (a; 2 + 2) Vl — x\ 

dx 

4. dy = sin -1 a; V = i (sin -1 a;) 2 . 

(1 — x 2 )* 

162. Integration of dy — =- a - 

* « + & cos 



= /*— 



dd 



b cos 6 

dd 



a (cos 2 ^ -f sin 2 -J + £ (cos 2 ^ — sin 2 - j 



-/ 



EXAMPLES. 

eld 



-/ 



(a + b) cos 2 - 4- (a — b) sin 2 ~ 
sec' 5 dd 

a 

a -{- b + (a — b) tan 2 jr 



-»/ 



^•tan 



^ 



a 4- b + (a — b) tan 2 



When ay b, 



tan -1 



.Va 4- #/ ^ J 



Va? - 6 2 

(by Ex. 3, Art. 132.) 

"When a < b, we have, from (1), 
£0 







/ ^_ = 2 r 

t/ a 4- 6 cos J 



d- tan - 



311 



(1) 



J 4- a — (b — a) tan 2 







log. 



a/# 4- a 4- Vb — a tan - 



Vb 2 — a 2 $ 

Vb 4- a — Vb — a tan - 

(by Ex. 5, Art. 137). 

dd 



The integral of 

b « 4- b sin 

to be 



can be found in like manner 



. 6 1 

a sm - 4- * cos - 

tan -1 - -tt , when a > 5 ; 






312 EXAMPLES. 

x a tan | + b - (6 2 - « 2 )* 
and, = f log , 

v J a tan g + b + (6 s . — a«)* 

when a <Cb. 

There are other forms which can be integrated by the 
application of the formula for integration by parts (Art. 
147). Those which we have given are among the most 
important, and which occur the most frequently in the 
practical applications of the Calculus. The student who 
has studied the preceding formulae carefully should find no 
difficulty in applying the methods to the solution of any 
expression that he may meet with, that is not too compli- 
cated. 

The most suitable method of integration in every case 
can be arrived at only after considerable practice and famil- 
iarity with the processes of integration. 

EXAM PLES. 

.1. dy = -~^=. y=^*(* + 8)(l -*»)*. 

2. dy = 



3. dy 



VI — ^ 

lx* 1-3 \ /- r 1-3 . i 

y = - \i + 2.1*7 Vl - * + 2i sm *• 

x 7 dx 



4. dy = 



Vl - ^ 

/afi 1-6 . 1-4-6 , 1.2-4.6\ /T 

xHx 



V 'a + bx 2 
1 /. . lax* , 8a 2 \ ? ■ " , ^ 



EXAMPLES. 



313 



fj — _ c ^ x 

£ 7 Vl — # 2 



--( 



_L *' 5 J- 8 * 5 

6z 6 + 4- 6z 4 + 2 



SM 



1-3-5, 1 + V1 — a 3 
-3^6 ,0 « * 



6. <fy = 



efo 



aVl 4 x 

V'i+x 



- i log 



/ a/1 + a; - l \ 
Wl + a; + 1/ 



(See Ex. 1, Art. 142.) 

7. r??/ = (« 2 — aF)i dx. 

y = }x (a 2 - rt 2 )t + -^ a 2 z (rt 2 - 3*)$ 

, 5-3 ,,4 , 5-3 . ,a; 

8. % = z 3 (l + z 2 ) 1 ^'. y = ^Vlh (1 4 * 2 )i 

9. dy = (1 -x^dx. 

y = \x (1 - z 2 )t 4 fz (1 - z 2 )* 4 f sin" 1 ^. 

10. dy = 



11. <fy = 



(0 _|_ ^2)| 

_ / 1_ 2\ i 

y== \a + bx* J 3ax /a 

dx 



SaVa 4 bx 2 



(a 4 &z 2 )* 



14 



btf? 3ff(a + W) ^ 3« 2 J 5aA /S4^F 



314 

12. dy 

y 

13. dy 

y 

14. dy 

15. dy 

16. dy 

17. dy 

y 

18. dy 

19. dy 

20. dy 

21. <fy 

y 



EXA3IPLES. 



Mx 



(1 - Z 3 )t 

X s — OX 



I sin -1 x. 



*Vl — z 2 

V2ar» — a: 2 

/a: 3 7.^2 7-5z 2 , 7.5.3 A , 

" ll +4T3 a + 4^2 a + 4^2«7 V2«*-< 



log £ efc. 

x 2 log 2 as c?^. 

dx 
x log a? 

# log 3 x dx. 



-log 3 as — — log 2 z 



+ ^~^^7 a 4 vers" 1 — 
4- 3-2 a 

y = a; (logs— 1). 

y = |a; 3 (log 2 x — I log as + |). 

y = log (log »). 



3a; 2 



log x — fa; 2 . 



a: 3 log x dx. 

dx 
x log 2 X 

log 2 a; das 

1 * 

as 8 

# 4 da; 



a; 4 , a; 4 

4 l0 ^-16- 



y = - 



log X 



^(log 2 a> — 4 log a; + 8). 



x% 



log 3 X 



x 5 5x 5 25 

2 log 2 a; 2 log a; + 2 



log (log x 5 ) 



+ log z 5 + ^- 2 log 2 a* + ~ log 3 a* |. 



EXAMPLES, 315 

22. dy — e*Mx. y = e x (x 4 —4:xz + 12z 2 — 24z + 24). 

a x I 1 \ 

23. ofy = za^a;. ?/ = ; [x — i )• 

^ * log a \ log «/ 

24. % = zWz. y = e x (x 2 — 2x + 2). 

25. dy = ^. y = - <r*(a? + 2z + 2). 

26. dy = -^- 

y = ~ 2^ + x ga ^ + ~%~ \ g X + g a ' X 

+ ilog2«.| +etc). 

27. <7y = ( %7+ i X ' y = log(e* + e-*). 

28. ^«/ = e e ' e x dx. y = e e *. 

e x xdx e? 

30. ^ = L__J_. Jr.^'ir+i)- 

[Put (1 + x) — z ; then jb = z — 1, tf» = *7z, etc.] 

c . sin 5 (9 ^/0 ,. . 

31. dw = 5-^— (Art, lo6.) 

* cos 2 v ' 

y = sec + 2 cos — ■£ cos 3 0. 

32. dy = sin* cos 3 d0. // = f sin* — f sin* 0. 

no , sin 3 dd - s „ rt i _ 

33. ^ = j # = f cos* — 2 cos* 0. 

cos*0 

34. rfy = 5^i?. 2, = 3sinie-fsinte. 



sin*0 



316 EXAMPLES. 

7 sin s 0^0 
3o. dy = 



cos 2 

y = — ^ (sin 4 + 4 sin 2 — 8). 

v 3 cos v ' 



36. dy = - 



sin 4 cos 2 

1 4 cos 8 cos 



* cos sm 3 3 sin 3 3 sin 

37. dy = — x cW , . y = 2 tan* (9(1 + | tan 2 (9). 

sin* cos* 

00 sin* (9 t?0 _ s 

38. ^ = g ?/ = f tan^ e. 

cos* 

39. <fy = . A f ., . « = - 8 cot 20 - | cot 3 20. 

* sin 4 cos 4 * 3 

40. dy = sin 4 cos 4 tf0. (Art. 157.) 

= — i?(cos 3 + icos0) - ^,- (sm 3 + |sin0) 

+ tI^- 

41. c?# = -r— „-:• # = sec + log tan -• 

y sm cos 2 J & 2 

42. #w = - — E ~ 

^ sm cos 4 

1 1,0 

V = n — t^ H ^ + log tan -• 

* 3 cos 3 cos 6 2 

43. ^ = sin 8 cos 6 tf0. 

QQg7 Q 

y = - -jj- (sin'0 +TVsm^ + ^sin 3 e+Asin 6) 



EXAMPLES. 317 

44. dy = sin 4 dd. (Art. 158.) 

y = Jz sin 40 — £ sin 20 + §0. 

45. dy — cos 4 tf0. 

y = gV sin 4(9 + i sin 20 + f 0. 

46. dy = sm & ddd. 

y — -gL (— -i sin 6(9 + | sin 40 — - 1 / sin 20 -f 100). 

47. d?/ = a, 4 sin x dx. (Art. 159.) 

y = —x* cos z + 4a; 3 sin x -f l&z 2 cos * 

— 24a; sin a; — 24 cos x. 

48. <fy = e™ sin 2 a; ds. (Art. 160.) 

e ax sin a; . 2e ai 

«/ = —7— — r- (a sin jc — 2 cos a) + 



4 + « 2 v > ' a (4 + a 2 ) 

49. tZy == e ax sin 3 a? dx. 

y — T \e x (sin 3 x + 3 cos 3 # + 3 sin a — 6 cos #). 

a sin kx -\- h cos &e 



oO. dy = e _<w sin kx dx. y = 

K1 7 a^cfa; sin -1 a; , . - . 

5L ^ = —77=^' ( Art 161 ') 
V 1 — x 2 



(a 2 + & 2 ) e a 



x^dx 
Put efo; = , - and u = sin -1 .?:; then 

Vl - & 
v = - i (a; 2 + 2) Vr 1 "^ (by Ex. 1), etc. 

3/ = -i(* 3 + ^) VT^- sin-ia + |* + |z. 

52. <7y = — - sin -1 a. 

Vl - S2 . 

y = - i (*+#) v / T = ^+ A- sin" 1 a: + ^+ A^ 2 . 



318 EXAMPLES. 

53 - *«rfei- ( Art - 16a -> 

x^dx 
54. d# = —==- (See Formula 43, p. 345.) 

y = ^tl(2-2x + x^ 

+ i log [x — 1 + 2 (2 - 2z + z2)i]. 



CHAPTER V 

INTEGRATION BY SERIES SUCCESSIVE INTEGRA- 
TION INTEGRATION OF FUNCTIONS OF TWO 

VARIABLES — AND DEFINITE INTEGRALS. 

163. Integration by Series.— The number of differ- 
ential expressions which can be integrated in finite terms is 
very small ; the great majority of differentials can be inte- 
grated only by the aid of infinite series. When a differen- 
tial can be developed into an infinite series, each term may 
be integrated separately. If the result is a converging 
series, the value of the integral may be found with sufficient 
accuracy for practical purposes by summing a finite number 
of terms ; and sometimes the law of the series is such that 
its exact value can be found, even though the series is infi- 
nite. This method is not only a last resort when the 
methods of exact integration fail, but it may often be em° 
ployed with advantage when an exact integration would 
lead to a function of complicated form ; and the two methods 
may be used together to discover the form of the developed 
integral. 

EXAMPLES. 

dx 
1. Integrate dy = — - — in a series. 
ct ~\~ x 

By division, 

1 

a + x 

r dx rn x x* x* .v. 

• V = / — i — = / » ir -* ■ 1 + etc. ) dx 

* J a + x V \ a a 2 a s a* ) 



320 EXAMPLES. 



_ X x % X B #* 

But /Vfa = lo S (« + *)• [ Art - 130 > ( 4 )0 

Oj /-y;2 /v3 /v4 

•'• 10g(a + a;) = «-^ + 3-S-^ + et0 - 
2. % = a£ (1 — x 2 ) i dx. 
Expanding (1 — x 2 )^ by the Binomial Theorem, we haye 

., /y>% sy<i /y>6 K/y>8 

(l-^ = l-|-|-^-g--etc. 



f i /, a* a 4 cc« 5z 8 , \ 



dx. 



= p 2 — \x* — -hpo' 2 ' — tjb« *~ — Tirw x ^' — etc. 



7 67$ 

3 - * = r+*' 



/r?^ = tan_la; [Art m > (16)] 



+ a 2 

/>>3 /v»5 /y? o*9 

*t/ *>t/ %JU %Aj . 

4. dy — r =• 

Vl + & 

y-» fly, 

— == = log (x + Vl + a?) 
V 1 4* #* 

[Art. 144, (1)] 

j£ 3 3a; 5 3-5^ 

~ * ~ 2-3 + 2^5 ~* 2747677 + etC * 

5. rfy = x^ (x — l)t dx. 

y = fyl + ±x* + ^aT* — irfr^ — etc. 



SUCCESSIVE IXTEGRATTOy. 321 

164. Successive Integration.— By applying the rules 
previously demonstrated for integration, we may obtain 
the original function from which second, third or n th dif- 
ferentials, containing a single variable, may have been 
derived. 

If the second derivative -~ 2 = X be given, when X is 

cix 

any function of x, two successive integrations will be 

required to determine the original function y in terms of x. 

Thus, multiplying by dx, we have 

P = Xta; 
dx 

or d.(^j = Xdx. 

Integrating, we get 

Multiplying again by dx and integrating, we get 
y = J X x dx + fc x dx = X + C t x + C 2 . 

d 3 u 
Similarly, if we had y| = X, three successive integra- 
tions would give 

x 2 
y = X 3 + C x ^ + C& 4- C z , and so on. 

Generally, let there be the n th differential 

p = x. 

dx n 



&)?**> 



322 SUCCESSIVE INTEGRATION. 

hence, by integrating we have 

jpf=7>* = *■ + ft 

Again, we get from this last equation, 

and by integrating, 

-^=2 = X 2 + C x x + (7 2 . 

Also from this we obtain 

d \-j-^~i) = X 2 dx + C x x dx -f C 2 dx, 

and integrating, 

d n ~ 3 y ' _ ■ x % „ 

And continuing the process we get, after n integrations, 

f U d n y = f Xdx" 

x n ~^ x n ~^ 

= Xn + Gi L2-3... (^-1) + ft 1.2.3...(w-2) 

+ . . . . cu* + a. (i) 



The symbol / Xdx n is called the n th integral of Xdx n , 

and denotes that n successive integrations are required. 
The first term X n of the second member is the n^ integral 
of Xdx n , without the arbitrary constants ; the remaining 
part of the series is the result of introducing at each 
integration, an arbitrary constant. 



DEVELOPMENT OF INTEGRALS. 323 

/» 
Xdx n into 

a Series. — By Maclaurin's theorem, we have 
f' l Xdx n = (f n Xdx n> ) + (f^Xdx^Y ? 

/ /»n— 2 \ ^.2 

+ (A*) 1.2-3 .*>- -!) + W i^fei 

+ V&/1.2... (* + 1) 

td?X\ x n+ * 

+ W/l^-.l^MJ + etC - (1) 

in which the brackets 

(/x<fe»), (/""W-i) .... (/*&), 

are the arbitrary constants 

G„, C^_!, . . . . 6 1? 

for that is what these expressions become respectively, 
when x = 0. 

By Maclaurin's theorem, we have 

v >_- idX\x /d?X\x* td*X\ x* 

which may be converted into (1) by substituting for 
x°, x\ x 2 , X s , etc., in (2), the quantities 

x n x n+l x n+ * 

-, etc., 



1.2...^' 2-3... (» + !)' 34...(n + 2)' 



* Since — f Xdxn - j Xdx n ~ 



1. 



324 EXAMPLES. 

and prefixing the terms containing the arbitrary constants 
as above shown, viz., 



1' ~»-*i.2> ' W1 1.2.3...(» — 1) 

(See Lacroix, Calcul Integral, Vol. II, pp. 154 and 155.) 



example s. 



1. Develop / 

J Vl-x 2 

Here X = (1 — x*)-* 



,o 1-3 . 1-3-5 . 
= 1 + ¥ 2 + ^^ 4 + 2^76^ + etc * 

Substituting in this series for x°, x 2 , x*, x*, etc., the 
quantities 

X* X 6 X 8 X 10 



1.2-3-4' 3.4.5.6' 5.6.7.8' 7-8. 9- 10 
and prefixing 

° 4 ' ° 3 1' ° 2 ]T3' C, I^3' 
we get 

A£t= = c 4 + c3 + a =4 + g, 

** a/i _ ^ 1 1-2 



, etc. ; 



VI r^ ~* ' ~ 3 1 ' "'1.2 ' ^1-2.3 

z 4 z 6 l-3z 8 



+ t-«-o-^-+ 



1.2.3.4 ' 2.3.4.5.6 ' 2.4.5.6.7.8 

1. 3- 5a; 10 

+ 2.4.6.7.8.9.10 + etC< 

2. Integrate d 3 ?/ .= 6« dx 5 . 

Dividing by dx 2 we have 

— I = 6# da, or c? ( y^l = 6a dx. 



EXAMPLES. 325 



/<ffl> -/*■*> 



or ^=6az+ft. 

Multiplying by dx and integrating again, we have 

cl l = Sax 2 + C,x + ft. 
ax 

Multiplying again by dx and integrating, we have 
y .= a& + ft - -f fts -f ft. 

3. Integrate cT 2 ^ = sin z cos 2 # efo 2 . 
Put sin x = z ; 

.-. dz = cos x dx, 
and dz 2 = cos 2 # dx 2 ; 

.\ d 2 y = z dz 2 ; 
from which we get 

y = - + ft* + ft ; 

sin 3 # ^ . _ 

/. y = — g (- ft sin a; + ft. 

4. Integrate d 3 y = ax 2 dx\ 



5. Integrate fZ 3 y = 2#- 3 d# 8 . 



Cz 2 
y = log a + — - + ftz + ft. 



6. Integrate <Z 4 # == cos x dx*. 



fta? Ox 2 „ 
y = cos x + -1- + -|1 + ftz + ft. 



6*!b INTEGRATION OF FUNCTIONS. 

166. Integration of Functions of Two or More 
Variables. — Differential functions of two or more vari- 
ables are either partial or total (Art. 80). When partial, 
they are obtained from the original function either by 
differentiating with respect to one variable only, or by 
differentiating first with respect to one variable, regarding 
the others as constant; then the result differentiated with 
respect to a second variable, regarding the rest as constant, 
and so on (Art. 83). For example, 

— - fix y) 

and £ % Ty= f{X ' y) 

are differential functions of the first and second kinds 
respectively, in which u is a function of the independent 
variables x and y. From the manner in which the 
expression cPu „ , x 

was obtained (Art. 83), it is evident that the value of u 
may be found by integrating twice with respect to x, as in 
Art. 165, regarding y as constant; care being taken, at 
each integration, to add an arbitrary function of y, instead 
of a constant. 

d 2 u 

167. Integration of dydx =f( 3C 9 V)- 

This equation may be written 

clu 
It is evident that -7- must be a function such that if we 
ax 

differentiate it with respect to y, regarding x as constant, 

the result will be / (x, y). 

Therefore we may write 

-£. =f(%>y)(¥ 



INTEGRATION OF PARTIAL DIFFERENTIALS. 327 

Here, also, it is evident that u must be such a function 
that if we differentiate it with respect to x, regarding y as 
constant, the result will be the function 



ff(x,y)dy. 
Hence, U = J Jf( x >y)dy \dx. 



Therefore, we first integrate with respect to y, regarding 
x as constant,* and then integrate the result with respect to 
x, regarding y as constant, * which is exactly reversing the 
process of differentiation. (Art. 83.) 

The above expression for u may be abbreviated into 

J J f (*» V) d V dx or J J f («, y) dx dy. 

We shall use the latter form ; \ that is, when we perform 
the y-integration before the ^-integration, we shall write dy 
to the right of dx. 

It is immaterial whether we first integrate with respect to 
y and then with respect to x, or first with respect to x and 
then with respect to y. (See Art. 84.) 

In integrating with respect to y, care must be taken to 
add an arbitrary function of x, and in integrating with 
respect to x to add an arbitrary function of y. 

In a similar manner, it may be shown that to find the 
value of u in the equation 

d s u , t . 

- — - — r = fix, y, z), 
dx dy dz ' v a ' ' 

we may write it 

u — J j J f(x, y, z) dx dy dz, 

* Called the y-integration and cc-integration, respectively, 
t On this point of notation writers are not qnite uniform. See Todhunter's Cal., 
p. 78 ; also Price's Cal.. Vol. n, p. 281. 



328 • EXAMPLES. 

which means that we first integrate with respect to z, regard- 
ing x and y as constant; then this result with respect to y, 
regarding x and z as constant ; then this last result with 
respect to x, regarding y and z as constant, adding with the 
^-integration arbitrary functions of x and y, with the 
^/-integration arbitrary functions of x and z, and with the 
^-integration arbitrary functions of y and z. (See Lacroix, 
Calcul Integral, Vol. II, p. 206.) 

EXAMPLES. 

1. Integrate d?u = bx 2 ydx 2 
Here dl-j-J = bx 2 ydx. 

du = \~bx z ydx + / (y) dx. 
.-. u =■- -^bx^y + f(y) x + <f> (y). 

2. Integrate d 2 u = 2x 2 ydxdy a 

Here t?( — ) = 2x 2 ydy. 

•"• £ = f^yty = x2 y 2 + w- 

du = x 2 y 2 dx -f (a;) t?o?. 
.% h = ^ 3 j/ 2 + /0 (a;) <fc +/(y). 

3. Integrate d 2 u = 3xy z dxdy. 

u = %xy + /(/> (a;) dx +f(y). 

4. Integrate i? 2 w = ax z y 2 dxdy. 

u = j% X V + f$ ( x ) dx +f(y)- 



INTEGRATION OF TOTAL DIFFERENTIALS. 329 

168. Integration of Total Differentials of the First 
Order. 

If u = f(x, y), 

we have (Art. 81), 

, du . du 7 
du = Tx dx+ dy dy > 

du du 

in which -=- dx and -j- dy are the partial differentials of u ; 



also, we havt 


> (Art. 84), 

d?u d*u 
dx dy ~ dy dx 9 




or 


d ldu\ d Idu\ 
dy \dx) ~ dx \dy) 


(1) 


Therefore, 


if an expression of the form 






du = Pdx + Qdy 


(2) 


be a total differential of u, we must have 






du p du n 

dx~ - r ' dy~ V ; 





and hence, from (1), we must have the condition 

dP_dQ 

dy ~ dx > {6) 

which is called Euler's Criterion of Integr ability. When 
this is satisfied, (2) is the differential of a function of x and 
y, and we shall obtain the function itself by integrating 
either term ; thus, 

u = /Pdx+f{y), (4) 

in which f(y) must be determined so as to satisfy the con- 
dition 



330 EXAMPLES. 

Remark. — Since the differential with respect to x of every term of 
w which involves x must contain dx, therefore the integral of Pdx will 
give all the terms of u which involve x. The differential with respect 
to y of those terms of u which involve y and not x, will be found only 
in the expression Qdy. Hence, if we integrate those terms of Qdy 
which do not involve x, we shall have the terms of u which involve y 
only. This will be the value of f(y), which added with an arbitrary 
constant to fPdx will give the entire integral. Of course, if every 
term of the given differential contain x or dx, f(y) will be constant. 
(See Church's Calculus, p. 274.) 

EXAMPLES. 

1. du = 4:X B y B dx + 3x*y 2 dy. 

Here P = kt?y\ Q = 3xhf. 

.-. ^- = 12zy and ^ = 12x*y 2 . 

Therefore (3) is satisfied, and since each term contains x 
or dx, we have from (4), 

u = f4:X 5 y 3 dx = x*y z -f- C. 

2. dti, = j+(2y-^di>. 

(3) is satisfied, therefore from (4) we have 

Since the term 2ydy does not contain x, we must have, 
from the above Eemark. f(y) = f2ydy = y 2 , which must 

x 
be added to -, giving for the entire integral, 

y y 

3. du = ydx 4- xdy. u = xy + C, 

4. du = (Qxy — y 2 ) dx + (3a 2 — 2xy) dy. 

u = 3x 2 y — y 2 x + C. 



DEFINITE INTEGRALS. c31 

5. du = (2<2rcy — 3fe 2 y) <£z -f (ax 2 — &Z 3 ) <??/. 

u = ax 2 y — J^rc 3 + C. 

The limits of this work preclude us from going further in 
this most interesting branch of the Calculus. The student 
who wishes to pursue the subject further is referred to 
Gregory's Examples; Price's Calculus, Vol. II; Lacroix's 
Calcul Integral. Vol. II ; and Boole's Differential Equations, 
where the subject is specially investigated. 

169. Definite Integrals. — It was shown in Art. 130 
that, to complete each integral, an arbitrary constant C 
must be added. While the value of this constant C remains 
unknown, the integral expression is called an indefinite in- 
tegral ; such are all the integrals that have been found by 
the methods hitherto explained. 

When two different values of the variable have been sub- 
stituted in the indefinite integral, and the difference between 
the two results is taken, the integral is said to be taken 
between limits. 

In the application of the Calculus to the solution of real 
problems, the nature of the question will always require 
that the integral be taken between given limits. When an 
integral is taken between limits, it is called a definite 
integral.* 

The symbol for a definite integral is 

>b 
f(x) dx, 



/. 



which means that the expression f(x) dx is first to be inte- 
grated ; then in this result b and a are to be substituted 
successively for x, and the latter result is to be subtracted 
from the former ; b and a are called the limits of integra- 
tion, the former being the superior, and the latter the 
inferior limit. Whatever may be the value of the integral 

* In the Integral Calculus, it is often the most difficult part of the work to pass 
from the indefinite to the definite integral. 



332 DEFINITE INTEGRALS. 

at the inferior limit, that value is included in the value of 
the integral up to the superior limit. Hence, to find the 
integral between the limits, take the difference between the 
values of the integral at the limits. 

In the preceding we assume that the function is continu- 
ous between the limits a and b, i. e., that it does not become 
imaginary or infinite for any value of x between a and b. 

Suppose u to be a function of x represented by the equa- 
tion 

u =f(x); 

then du = f (x) dx. 

Now if we wish the integral between the limits a and b, 
we have 

u= f b f'(*)dx=f(b)-f(a). 

If there is anything in the nature of the problem under 
consideration from which we can know the value of the 
integral for a particular value of the variable, the constant 
C can be found by substituting this value in the indefinite 
integral. Thus, if we have 

du — (abx — bx 2 )* (ab — 2bx) dx, 

and know that the integral must reduce to m when x = a, 
we can find the definite integral as follows: 

Integrating by known rules, we have 

u — \ {abx — bx*)% + C, 

which is the indefinite integral ; and since u = m when 
x = a, we have 

m = -f C; .-. C = m, 

which substituted in the value of u gives 
u == | (abx — bx 2 )i> + m. 



EXAMPLES. 



333 



EXAMPLES. 

1. Find the definite integral of u ■— (1 -f- %ax)%dz 9 on 
the hypothesis that u = when x = 0. 

The indefinite integral is 

Since when # = 0, u = 0, we have 

° = Wa +C > •'■° = -Wa' 
which substituted in the indefinite integral, gives 

w = 4 (1 + ffl ' r)i - 4 
for the definite integral required. 

2. Integrate du = 6x 2 dx between the limits 3 and 0. 

* 



Here 



Jo 



6x*dx = 



2z* 



= 54. 



a -f 1 



3. u = I x n dx z= 

Jo \_n + 1 

4. u = f*e-*dx = [~— e- x ~r= — (0 — 1) = 1. 

5. u = / - r - — 5 = - tan -1 - = — • 

Jo a 2 + £ 2 a L. rt Jo ^ a 



6. w 



Jo a 



a dx 



+ x* 



tan 






* This notation signifies that the integral is to be taken between the limits 3 
and 0. 



334 



CHANGE OF LIMITS. 



/°° dx If. ,x" 

-5— — 5 = - tan -1 - 



= - Ttaii -1 00 — tan -1 (— 00 )1 = -• 






^ 



V# 2 — # 2 



sin -1 



Remark. — It should be observed here that the value of the infini- 
tesimal element corresponding to the superior limit is excluded, while 
that corresponding to the inferior limit is included in the definite in- 
tegral ; for, were this not the case, as — becomes equal to 00 

when x = a, the integral of Ex. 8 between the limits a and would 
not be correct ; but as the limit a, being the superior limit in Ex. 8, 
and that which renders infinite the infinitesimal element, is not 
included, the definite integral is correct. (See Price's Calculus, 
Vol. IT, p. 89.) 

9. u = ^(a 2 — x*)?dx. (See Ex. 4, Art. 151.1 
^0 



10. u 



|K-^)i + |%in-^ ' 



= f 1 f^L_ . (See Ex. 7, Art. 151.) 

t/ Vl — z 2 

_ 1-3.5- 7T 

— 2.4.6.2* 



a'TT 



11. 



/ sin 7 x cos 4 x dx 
Jo 



42 



3.5.7.11 



170. Change of Limits. — It is not necessary that the 
increment dx should be regarded as positive, for we may 
consider x as decreasing by infinitesimal elements, as well as 
increasing. Therefore, we have 



CHANGE OF LIMITS. 335 

/> (x) clx = <t> (a) - <l> (5) = - [0 (b) - (a)] 

^b 

= — I 0' (#) $£. 

That is, if we interchange the limits, ive change the 
sign of the definite integral. 

Also, it is obvious from the nature of integration (Art. 129), that 

pc r»b r>c 

I <$> (.r) dx = f 6 (x) dx + I (p (x) dx, 
J a J a Jb 

and so on. Hence, 

fin fi\n p\n n\n 

I cos x dx = I cos x dx + / cos x dx + I cos x dx 

t/0 t/0 J\n J\n 

+ I cos x dx 

Jfr 

r*\n r*n 

= / cos x dx + I cos x dx = 0. 

t/0 Jin 

Also, r a f{x)dx= C f'(x)dx + Hf'Wdx. (1) 

J —a J -a Jo 

Let x — — x ; then dx = — dx, and the limits and — a become 
and + a ; therefore we have 

/0 p0 pa 

f'(x)dx=- / f'(-x)dx= / f'{-x)dx, 
a J a t/0 

which in (1) gives 

pa pa pa 

/ f'(x)dx = / f'(-x)dx+ / f'{x)dx 
J -a JO JO 

= f a [f'(-x)dx+f'(x)dx]. (2) 

t/0 

Now if /' (—x) = -/' (x), (2) becomes, 



pa 

/ f'{x)d, 
J -a 



v = 0. 



33 G EXAMPLES. 

But if /' (—a?) =/' (a>), we have 

J* f (x) dx = rjf (x) dx, (3) 

Xa pa 

f (x) dx = 2 I f {x) dx. (4) 

The following are examples of these principles. 

/in 
cos x dx. 
in 

Here /' (— x) =f (x) = cos x. 

/*h* pi* 

cos x dx = 2 / cos x dx = 2. 
-in Jo 

2. u = / "smx dx = 0. [Since/' (— x) = —f (x).~] 

pa < pa , -rrrfi 

3. u = / (a*-x*)idx = 2 (a? — xrfdx = -^- 

J -a Jo Z 

piT p\tt pit 

4. w= / smx dx = / smxdx+ / sinxdx 

Jo Jo J\n 

p\n 

= 2 / sin x dx. 
Jo 

Since / sin x dx = I sin x dx \ = 2. 

/J7T / i l 7r / j7r 

5. u = / cos x dx = I cos x dx + / cos # <fa? = 0. 

Since / cosx dx = — / cos # <£r. 

z 11 # 4 d:r _ r 1 x*dx . : 

6. U = —=r = 2 / — - = i'fTT. 

(See Ex. 2, Art. 162.) 



EXAMPLES. 



337 



1. Integrate du = 



exam ple s. 
dx 



Va 2 — a' 2 



by series. 



x x 

+ 



1-3.T 5 1-3. 5s 7 

t- »,-..« + ^-r^r-^ + etc. 



a ' L2.3a 3 ' 2.4-oa 5 ' 2-4.6.7a 7 



But/ 



«?» 



V^ 2 



= sin- 1 - (by Ex. 14 of Art. 131) ; 



therefore, 



sin" 



x x X s 



2. Integrate dfo = 



1.3a? 1.3.5a? 

+ TT- a o » * + etc « 



a a ' 2-3a 3 ' 2-4.5« 5 ' 2.4.6-7« 7 
dx 



VI 4- ** 

_ a- _ I^t 5 1.3a? 1.3-oa: 18 
W ~~ 1 2-5" + 2.4.9 2-4. 603 



+ etc. 



3. Integrate du — 

Vl - x> 

By the Binomial Theorem, 



(1 _ <fo2)i. 



Vl - e 2 a; 2 = 1 _ i*gP 4- ll^ 4 _ ll^ 6 ^ 6 -f 



2-4 2-4-6 



<?.r 



Multiplying by — ~=L=== , and integrating each term sep- 
V 1 — a; 2 

arately (see Ex. 1, Art. 151), we have 



u 



f 



dx 



VT 



(1 _ e 2 X 2)i 



sin -1 x + 



+ 



1^2 



2-4 



~ Vl — a; 2 — \ sin -1 a; 
"/a? i.a.«\ 7l 1 1-3 . _ t "I 



15 



338 



EXAMPLES. 



1.3- e 6 /«« l-faj 1.3-5a?\ /=— - 
+ 2-4.6 L\6 4-6 + 2.4.6/ V 



& 



1^3-5 

2-4^6 



sin' 



-^l 



+ etc. 



2 



4. Given #y = - 3 ^ 3 , to find y. (Art. 164.) 

2/ = logo; + \C x x 2 + ftz + ft. 

5. ## = — x'^dxK 

y = 1 log a + |ft:B 3 + ifta 2 + ftz + <7 4 . 

6. ^ = #W. y = yf^i 4- ift^ 2 + ftz + ft. 

7. ^ = &*fo* y = T V^ 4 + C x x + ft. 

8. d 2 y = cos a; sin 2 x dx 2 . 

y = 1 cos 3 x + C x cos sb 4- ft. 

9. rf 4 y = cos x dxK 

y = COS SB -f ^ftiB 3 -f J-CgSB 2 + C 3 x + ft. 

10. d 3 y = e*dx 3 . y = e* + iftz 2 + fta -f ft. 

11. #y = (1 + x*)-Ha*. (Art. 165.) 

/y>2 <y3 /v>4 

y=0 4 -f^+ft^ + ft^-3+^ i 



r^ 6 



+ 



l-3z 8 



2.3.4.5.6 ' 2.4.5.6.7.: 

1.3. ox™ 



2.4.6.7.8.9.10 



+ etc. 



12. d 2 u = ax^yWxdy. 

m = ^^if + f</>(x)dx +f(y). 

13. 6?% = (2a;?/ 2 + 9a 2 # + Sx*) dx + (2a% + 3sb 3 ) dy. 

u = x 2 y 2 + Zxhj + 2z 4 + 0. 



EXAMPLES. 339 

14. u = f a (a} - x*f~ dx = 2 fid* - x*)i dx. 

(See Art. 170.) 
u = h^ ^n. (See Ex. 7 of Art. 162.) 



15. u = / 



5-4-6 

2 « artafe 



V 2## — x 2 

7.5.3 . 

2^' 

2 5 



16. ^f^Hl-^dx^^^^ 

17. w = f 1 —^— (1 _ <&«)*. (See Ex. 3.) 

e/o Vl - ^ 

_ TT 1 2 7T 1.3 4 7T_ 1.3 2 .5e 6 7T 

W "2""4 e 2~2^T2 e 2 ~ 22.42.62 2 ~ etC * 

18. u= /*ri*L* (Art. 167.) 

We first perform the ^/-integration, regarding x as con- 
ant, and then the ^-integration. 

.-. u = f z R Uar^^Txdx. (Art. 132, Ex. 3.) 

Jq \_X Xjo 

- J 4 ** ~ 16 

/»a r*x ay 

19. w = / / / ^2 da; dy dz 

= J t J 9 t dxd y = J 8 &= 48- 

pa pa-x pa—x-y /*3 

20. u= / / dxdydz = £-. 

^0 ^0 ^0 « 



340 FORMULAE OF INTEGRATION. 

21. 



oft* pa (fix 

. u = I f r B dd dr = -^- • 

e/ ^o 2 

/|tt />2a cos 
/ r 5 dd dr == fa 4 *. 



/>2a cos 

22 



For the convenience of the student, the preceding for- 
mulae are summed up in the following table. 



TABLE OF INTEGRALS. 

CHAPTEE I. 

Elementaky Fokms. (Page 238.) 

1. J (dv + dy — dz) = v + y — z. (130)* 

ax n dx = -^— - • (131) 

n + 1 v ' 



Padx _ a 

J ~xF " " ~~ (n — 1) x 11 ' 1 ' 

Padx 

4. J — = a log a. 

5. / a x log adx = a x . 

6. / e x a# = e x . 

7. y cos xdx = sin #. 

8. / sec 2 xdx = tan a;. 

9. / sec a; tan x dx — sec #. 



* The arbitrary constant is understood. (See Art. 131.) 



FORMULAE OF INTEGRATION. 341 



-•^ P dx 1 . 1 bx ,„ rt *i 

10. / — = = - sir 1 - • (132) 



"•/ 



^r 1 . , 5a; 
— = T sm -1 - 

- w> b y 

f x 1 _ t bx 

a 2 + b 2 x 2 ~ ab a 



12. f — 



dx 1 , bx 

, — = - sec -1 — • 

x y/W — a 2a a 



dx 1 1 bx 

= t vers -1 — • 



13. f-_ 

J V2abx - b 2 x 2 *> <* 

14. / tan x dx = log sec x. 

15. / - — = loe- tan -%x. 

J sm x to * 

CHAPTER II. 
Bational Fractions. (Page 256.) 
Pf(x)dx _ P Adx . Bdx PLdx, „ 



f(x)d, 
- (x) 



P Adx P Bdx P Ldx 

-J j£psr+l <£=sp+ ■ • ■ J &zr a x (138) 

*/ (b (x) ~ J [(x ± a) 2 + b 2 ]' 1 
P (Cx + D)dx P {Kx + L)dx 

J [(x ± a) 2 + ¥¥~ l "*"•■• J (x± a) 2 + # 2 ' { } 

iq C—^L— — _tl l / g "*" ^ — 

• J a 2 - b 2 x 2 ~ 2ab g \a -bx)' ' 

OA P adx - ■ ~--Jx — a- 

20. / -= 5 = losr \ / -■ 

J x 2 — a 2 - \ x + a 



342 FORMULAE OF INTEGRATION. 

CHAPTER III. 
Irrational Functions. (Page 269.) 

[where z = (a + bx)?]. (142) 

99 r x 2n+ldx r (* 2 — a ) n dz 

' J (a , + hx i)\ ~ J ** > 

[where z = (a + W)*]. (143) 

23. /-—£?===. = log(l+x+ V^+bx~+x~A (144) 

24. y~ 



' — = — 2 tan" 1 A / £ _^ 

v« + &» — ^ 2 V ^ — « ' 



01 



[where a + bx — x 2 = (x — cc) ((3 — x). 

25. fx'\a-\-bx n )Ux=^f^ + <-»(p^) " dz,(U6) 
(where 2* = a + bx n ) ; 

(where £ tt z y = a + #af). 

26 . /•_*=„ A log (v«±^=V^\ . E? 6> (146) 



or 



. /L^*L_==1 log (&,+ V« 2 +^ 2 ). Ex. 16. 

J Va*+ Wx 2 & 

. /• £ == Ilogf^±^^Y Ex.17. 

J zV'a 2 + b 2 z 2 a \ bx / 

1, /a4-Va 2 4-^ 2 \ 

=^ lo n — bx—y 



FORMULAE OF INTEGRATION. 343 

29. f~~^= = a log (x + a + V&w.+ a?). Ex. 20. 

30. f (&+&)* dx 

= % (a 2 + z 2 )* + 1- 2 log [a + (a 2 + xrfl Ex. 35. 

CHAPTEE IV. 

Successive Beductiox. (Page 285.) 

31. / udv = uv — I vdu. (147) 

32. J'z m {a + hx tl Y dx (148) 

x m ^ ,+1 (a + bx n y +l —(m— n + 1) a fx m - n (a + bx n ydx 

. '-{ /^\ 

b (np + m + 1) 

33. Jx" m (a + bx n Y dx (149) 
x -M+-i ( a _f_ fof )r+i + #( m _ ^ _ n _ i ) r x - m+n {a + bx li ydx 

— « (m — 1) 

34. fx m {a + bx n Y dx (150) 

af** 1 (« + bx a )v + «^ / x m (a + bx" l ) p ~ l dx 
= — _ «£ ((7) 

np + m -\- 1 

35. y af 1 (« + &»»)-? dx (151) 

a Wi+1 (« + Jaf l )~p +1 — (m + ».+ 1 — rap) A," n (a + Jaf l )-* +1 (fe 

-.(Z>) 



*m ( 2? — 1 ) 
36. Jx m (a 2 — z 2 )-i dx 






344 FOBMULJB OF INTEGRATION. 



37. fx m {a 2 + x 2 )~? dx 

3 8 . r — *l_ 

d x m (a 2 — x 2 )? 

(a 2 — x 2 f~ m — 2 r dx 

~ (m — l)a 2 x m ~ l (m — l)a 2 J x m ~ 2 (a 2 — x 2 )? 

39. f{a 2 -x 2 fdx 

x (a 2 — x 2 f + na 2 f \a 2 — x 2 f~'dx 



40. 



A 



n -f 1 

dx 



41 



(a 2 — x 2 ) 2 

x n—S r dx 

(n-%)a 2 (a 2 -x 2 )?-' (n-2)a 2 J (tf—x*)?- 1 

x m dx * 



. r 

v V2ax — x 2 

x m ~ x /K - 9 (2m — l)a r x m ~ l dx 

— y2ax — x 2 + ^ — / — « 

" va + bx -\- ex 2 

af -1 Va + bx + ex 2 n — 1 a P x n ~Hx 



/: 



2w— 1 £ /* x n ~ l dx * 



t — 1 * /■ 

2w c«/ 



2?i C «/ y^ _|_ J T _[_ ^2 



* See Price's Calculus, Vol. II, p. 63. 



FQRMTFLJE OF INTEGRATION. 345 

43 . r **L . 

" Va + bx + ct 2 

(x 3b\ (3V> a\ r dx * 

:Va + bx + cx^-^)+ {&-Je)J V a + bz+c*? ' 



x dx 
V 'a + bx -f ex 2 



ya + to + ex 2 b 

c %c u Vet + bx + cz 2 

y « + &z + cz 2 a r_l 



/* dx 

J Va'+bx 



c 



B^-W+V^? + ? 



45. / of 1 log n &•<#£ 

t log' 1 sb / x m log*"" 1 zd'z. (152) 



m + 1 

46. /£* 

t/ loff"a; 



T m+i m -f 1 /* £ m $r 



^ + 



=1V lo^T (15d) 



(w — 1) log' 1-1 a; n — 1 1/ log" -1 a; 

«/ log 2 

= log (log 2) + log 2 + - 2 log 2 z + 2^3 log8 * + etc * 



47. * 



a'" x x n dx = -At-- ~ — / a mx x n - l dx. (154) 

m Jog a m Jog # «/ v ' 

ra x dx _ a x log a ra x dx 

' J x m " ~~ (^T~i) ^-1 + ~^ZT\J x m ~ 1 ' *■ ' 

50. J sm m 6 cos n 6 dd 

= — f(l — cos 2 ey- cos n (96? cos 0, (156) 

* See Price's Calculus, Vol. II, p. 63. 



346 FORMULJE OF INTEGRATION. 

(when m = 2r + 1) ; 
or = y tan» d (1 + tan 2 (9)'- 1 ^ tan B, 

(when m -\- n = — 2r) ; 
or = fx m (1 — x*f^dx, (when # =.sin 0). (157) 

51 . I x n cos «# dx 
= — ^-te sin ax + w cos ##) — ^— - / x n ~ 2 cos aa: c?a;. (159) 



ax nr\Q n — 1 



^^ /• 7 e ax cos" J x (a cos a; + n sin a:) 

52. / e ax cos" xdx = ^ r-^ '- 

+ — \ -^ / e ax cos"- 2 x dx. (160) 

#2 _l_ n 2 J v ' 

53. J sin" 1 xdx z=x sin" 1 a; + (1 — a; 2 )i (161) 
«/ # + b cos 

= vsb ^'[(srlfB (when a>J) - (163) 

= ^L= lQ [ -VgTa+V^ta n | -| 

\/W—a* \-\/bA-n.— \/h—aAsix\l\ 



Vb 2 —a 2 ^Vb + a— Vb— a tan 

, m £ , /**\ ^ 

^ v ' 1-2-3 ... n "*" \ dx/l-2-Z .. . (n + 1) 



CHAPTER VI 



LENGTHS OF CURVES. 

171. Length of Plane Curves referred to Rectan- 
gular Axes. — Let P and Q be two consecutive points on 
the curve AB, and let (x, y) be the 
point P ; let s denote the length of 
the curve AP measured from a fixed 
point A up to P. Then 

PQ = ds, PR = dx, RQ = dy, 

Therefore, from the right-angled 
triangle PRQ we have 



ds = v dx 2 + df ; 
hence, s = / y dx 1 -f dif = II 




1 + 



dif 



9 ch 



To apply this formula to any particular curve, we find 

dy 
the value of -j- in terms of x from the equation of the 

curve, and then by integration between proper limits s 
becomes known. 

The process of finding the length of an arc of a curve is 
called the rectification of the curve. 

It is evident that if y be considered the independent 
variable, we shall have 



= /(- +&+ 



The curves whose lengths can be obtained in finite terms 
are very limited in number. We proceed to consider some 
of the simplest applications : 



348 RECTIFICATION OF THE PARABOLA. 

172. The Parabola.— The equation of the parabola is 

y 2 = %v x ; 

hence, -~ = -• 

ax y 



r' p 



or 



= ^/^ 2 + l/ 2 )*^ (which, by Ex. 35, Art. 146) 



= ^_ti? + fio g(yW F+7) + a 



(i) 



If we estimate the arc from the vertex, then s = 0, 
y = 0, and we have 

0=|log^+C; .-. C= -flog*, 

which in (1) gives 

s - ^ + g log ^ ^— — j , (2) 

which is the length of the curve from the vertex to the 
point which has any ordinate y. If, for example, we wish 
to find the length of the curve between the vertex and one 
extremity of the latus-rectum, y = p, we substitute p for y 
in (2), and get 

5 =i^A/2+|log(l + V2) 
for the required length. 

We have here found the value of the constant C by the 
second method given in Art. 169. We might have found 
the definite integral at once by integrating between the 
limits and p, as explained in the first method of Art. 169, 
and as illustrated in the examples of that Article. Hence, 



RECTIFICATION OF THE CIRCLE. 349 

we need not take any notice of the constant (7, but write 
our result 

s = - f 1 \f + y*)%dy, (see Art, 169) 
and integrate between these limits. 

173. Semi-Cubical Parabola.* — The equation of this 
curve is of the form y 2 = ace?. (See Fig. 39. ) 

Hence, —^ = f ^/ax and -^ = %ax. 

CIX CIX" 

/. s = J (I -f- ±ax)? dx 

=-- £- a (l + iax)i + a 

If we wish to find the length of the curve from A to P, 
we must integrate between the limits and Sp (see Art. 
128, Ex. 9) ; hence, 

s = J^ (1 + i«x)i clx = I JL (l + j«r)l I 

= 2^[(1+¥«P) I -1] =P(3*-1), 

by substituting — — for a. (See Art, 125, Ex. 1. Compare 
Ex. 10, Art. 128.) 

174. The Circle.— From x 2 + if = r 2 , we have 

dy _ x 
dx~ ~7 

* This was the first curve which was rectified. The author was William Neil, 
who was led .to the discovery, ahout 1660, by a remark of Wallis, in his Arithmetica 
Iufinitorum. See Gregory's Examples, p. 420. 



350 RECTIFICATION OF THE ELLIPSE. 

Hence, for the length of a quadrant, we have (since the 
limits are and r), 

r r /. ■ x 2 \k , P r rdx 
s _ / (1 + -=) dx _ / — - 

tx~~\ r 

which involves a circular arc, the very quantity we wish to 
determine. The circle is therefore not a rectifiable curve ; 
but the above integral may be developed into a series, and 
an approximate result obtained. 
By Ex. 1, Art. 170, we have 

f (x ■ x s 1-Sx 5 1.3.5a' \~} r 

S = \_ r \r + W + 2^p + 8^6-fi* + etc '/J 

/, l 1.3 3-5 , \ 

1 , 1.3 3.5 

By taking a sufficient number of terms, reducing each to 
a decimal, and adding, we have n _ 3. 141592653589793 + . 
For the approximation usually employed in practice, n is 
taken as 3.1416, and for still ruder approximations as 3-J-. 

175. The Ellipse. —From if _ (1—e 2 ) (a 2 —x 2 ), we have 

dy _ n 2\ x - x ^^ ~~ e2 
_____ ( _ e) _ __ _ .____. 

To find the length of a quadrant, we must integrate be- 
tween the limits and a ; hence, 

P a I _ AxPti dx 



V- 



RECTIFICATION OF THE CYCLOID. 



351 



This integration cannot be effected in finite terms, but 
may be obtained by series. 

x 
Put -=z: then dx = adz. When x = a. 2 = 1, and 
a 

when $ = 0, z = ; therefore the above integral becomes 



dz 



8 = a (1-eW)?—^ 

^0 A/1 — 



1-3 



1-32-5 



= fl I I 1 " **'- a^ 6 * " srpisi.^ - etc -)' 

(by Ex. 17, Art. 170), which is the length of a quadrant of 
the ellipse whose semi-major axis is a and eccentricity e. 

176. The Cycloid.— From x= rvers -1 - — V2ry—y% 

we have 

dx _ y 

dy ~~ ^/2 r y — f 
.-. s = V'2r J (2r — y)~* dy 



r i 2r 



which is ^ the eycloidal arc; 
hence the whole arc of the cy- 
cloid is 8r or 4 times the diam- 
eter of the generating circle. 

If we integrate the above ex- 
pression between y and 2?*, we get 




Fig. 44. 
s = V2r / (2r — y)~? dy •= 2 (2r)* (2r — y)* 



= 2 V2r (2r — y) — arc BP. 



But BD = a/BA x BC = V'2r (2r - y) ; 

.-. arc BP = 2 times chord BD.* 



* This rectification was discovered by Wren. See Gregory's Examples, p. 421. 



352 



INVOLUTE OF A CIRCLE. 



177. The Catenary.— A catenary is the curve assumed 
by a perfectly flexible string, when 
its ends are fastened at two points, 
A and B, nearer together than the 
length of the string. Its equation is 



y 

Hence, 

dy 

dx 



(e« + e a ). 




W- e ~ l Y 



Fig. 45. 
ds = -(e° + e~~ a )dx. 



If s be measured from the lowest point V, to any point P 
(x, y) y we have 

178. The Involute of a Circle.— (See Art. 124.) Let 
be the centre of the circle, 
whose radius is r ; APR is a 
portion of the involute, T and 
T' are two consecutive points 
of the circle, P and Q two 
consecutive points of the in- 
volute, and <f> the angle ACT. 
Then TCT' = PTQ = dtf>, 
and PT = AT = rcf>. 

.-, ds = PQ = r0# ; F, 'S- 46 ' 

.-. s = rj*<f)d<l> — \r<p + C. 

If the curve be estimated from A, (7 = 0, and we have 
s = %r<j> 2 . 
For one circumference, = 2n ; .\ s = \r (2n)^ = 2rrr 2 . 
For n circumferences, (f> = 2nn; .-. s = %r(2nn) 2 = 2rn 2 n 2 . 




THE CARDIOIDE. 353 

179. Rectification in Polar Co-ordinates. — If the 

curve be referred to polar co-ordinates, we have (Art. 102), 

ds — vH& + dr 2 ] 
hence we get s = I (r 2 -j- ^j dd, 

s =J [l + ^j dr. 

180. The Spiral of Archimedes. — From r = ad, we 
have 

dd _ 1 



a J v ' 



2a + 2 g \ a /' 



(see Art. 172), from which it follows that the length of any 
arc of the Spiral of Archimedes, measured from the pole, is 
equal to that of a parabola measured from its vertex, r and 
a having the same numerical values as ij and m. 

181. The Cardioide. — The equation of this curve is 
r = a(l -f cos 6). 

Here — = — a sin 0, 

at? 

and hence s = y [a 2 (1 + cos 0) 2 + a 2 sin 2 d]$d$ 
— a J (2 -f.2 cosd)?dd . 

cos - dd — ±a sin - + C. 



354 



LENGTHS OF CURVES IN SPACE. 



If we estimate the arc s 
from the point A, for which 
= 0, we have 

s = 0; .'. (7=0. 

Making = n for the 
superior limit, we have 

s = la sin - = 4a, 




Fig. 47. 



which is the length of the arc ABO ; hence the whole 
perimeter is 8«. 

182. Lengths of Curves in Space. — The length of 
an infinitesimal element of a curve in space, whether plane 
or of double curvature, from the principles of Solid Geom- 
etry (see Anal. Geom., Art. 169) is easily seen to be 

Vdx 2 -f df + <& 3 -* 

Hence, if s denote the length of the curve, measured from 
some fixed point up to any point P (x, y, z), we have 



= J Vdx 2 + chf + d% % 



dz\n 



dx. 



If the equations of the curve are given in the form 
y=f{x) and z = ${x), 

Oil uX 

we may find the values of ~ and -=- in terms of x, and 
then by integration s is known in terms of x. 

* The student who wants further demonstration of this, is referred to Price's 
Cal., Vol. I, Art. 341, and Vol. II, Art. 164; De Morgan's Dif. and Integral Cal., 
p. 444; and Homershain Cox's Integral Cal., p. 95. 



EXAMPLES. 355 

183. The Intersection of Cycloidal and Parabolic 
Cylinders. — To find the length of the curve formed by the 
intersection of two right cylinders, of which one has its 
generating lines parallel to the axis of z and stands on a 
parabola in the plane of xy, and the other has its generating 
lines parallel to the axis of y and stands on a cycloid in the 
plane of xz, the equations of the curve of intersection being 



yi _ ^p X ^ z = a vers -1 - -f- V%ax — x 2 . 

ct 

tt fy _ lv a dz _ l%a — x > 

dx ~" v x dx ~" y x 9 

... A = (l+£ + £-l)*&=(, + *)*.£. 

Y X 

Estimating the curve from the origin to any point P, we 
have 

s = / (p 4. 2a)i ~ = 2(p + 2a)* V%. 

EXAMPLES. 

1. Rectify the hypocycloid whose equation is 

2,2 2 

£3 _|_ ^3 — £3. 

^4ws. The whole length of the curve is 6a. 

X 

2. Rectify the logarithmic curve y = le a ". 

^** ^ = « log — — - £- — + v« 2 + ?/ 2 + a 

a + ^2 + y2 

e x 4- 1 

3. Rectify the curve e y = ~~ between the limits 

x = 1 and x = 2. 

Ans. s = log (e + e _1 ). 



356 EXAMPLES. 

4. Kectify the evolute of the ellipse, its equation being 

©•+ © ! = ■• 

Put x = a cos 3 0, y — (3 sin 3 ; 

then dx = — 3a cos 2 6» sin 6 dQ, 

dy = 33 sin 2 d cos d0 ; 

.\ s = 3 /"^(a 3 cos 2 <9 + j8 2 sin 2 0)£ sin cos (9 d0 

3 /*** /«2 + 02 „2 _ £2 U 

= "4^0 (~T~ + T~ cos2 7 dcos2d 

~ «2 _ £2 ^ 

w 3 08 

therefore the whole length is 4-= -• 

a** — />* 

If (3 = a, this result becomes 6a, which agrees with that 
given in Ex. 1. (See Price's Calculus, Vol. II, p. 203.) 

5. Find the length of the arc of the parabola x^+y^ = a? 
between the co-ordinate axes. 

Put x = a cos 4 0, y = a sin 4 ; 

.-. 5 = 4a / (cos 4 -f sin 4 0)* sin cos dd 

= — -— f*(l + cos 2 2(9) d cos 20 

A/2 ^o 

= -« + 4i ,o s (^ + x )- 

V2 

6. Find the length, measured from the origin, of the 

curve / v\ 

x 2 = a 2 ll — e a ). 



Ans. s = a log ( '-) — x. 

& \a — x! 



EXAMPLES. 



!57 



7. Rectify the logarithmic spiral log r — 6 between the 



limits r Q and r x 



Ans. s = (1 + m 2 )* (r, — r ). 



8. If 100 yards of cord be wound in a single coil upon an 
upright post an inch in diameter, what time will it take a 
man to unwind it, by holding one end in his hand and 
traveling around the post so as to keep the cord continually 
tight, supposing he walks 4 miles per hour ; and what is the 
length of the path that the man walks over? 

Ans. Time = 51^ hours ; distance = 204 T 6 T miles. 

9. Find the length of the tractrix or equitangential 
curve. 

If AB is a curve such that PT, 
the length of the intercepted 
tangent between the point of 
contact and the axis of x, is 
always equal to OA, then the 
locus of P is the equitangential 
curve. 

Let P and Q. be two consecu- 
tive points on the curve ; let 
(x, y) be the point P, and OA = PT = a. Then 




Fig. 48. 



PQ 
PR 

cly 



y 



(the minus sign being taken since yis a decreasing function 
of s or x). 



Hence. 






J n. V 



logy 



= a log 



358 EXAMPLES. 

This example furnishes an instance of our being able to 
determine the length of a curve from a geometric property 
of the curve, without previously finding its equation. 

The equation of the tractrix may be found as follows : 

PR _ PM 

RQ "~ MT ' 



hence 



dx v^2_^' 

y Va 2 — y 2 



-/. 



py a 2 dy P 

J a J a 2 — u 2 Jo 



dy 

yfy 



Va 2 — y 2 ^° Va 2 — y 2 



= a log *±?f^* -(*-#. 

(See Ex. 17, Art. 146.) 

This curve is sometimes considered as generated by attaching one 
end of a string of constant length (= a) to a weight at A, and by 
moving the other end of the string along OX ; the weight is supposed 
to trace out the curve, and hence arises the name Tractrix or Tractory. 
This mode of generation is incorrect, unless we also suppose the fric- 
tion produced by traction to be infinitely great, so that the weight 
momentum which is caused by its motion may be instantly destroyed. 
Price's Calculus, Vol. I, p. 315. 

10. A fox started from a certain point and ran due east 
300 yards, when it was overtaken by a hound that started 
from a point 100 yards due north of the fox's starting-point, 
and ran directly towards the fox throughout the race. Find 
the length of the curve described by the hound, both having 
started at the same instant, and running with a uniform 
velocity. Ans. 354.1381 yards. 

This example, like the preceding, may be solved without finding 
the equation of the curve. 



EXAMPLES. 359 

11. Find the length of the helix, estimating it from the 
plane xy, its equations being 

x = a cos 0, y = a sin 0, z = c$. 

Ans. s — (a 2 + c 2 )*(p. 

12. Find the length, measured from = 0, of the curve 
which is represented by the equations 

x = (2a — b) sin — (a — b) sin 3 0, 
y = (2b — a) cos cf> — (b — a) cos 3 0. 
^s. 5 = \ (a + #) + | (« — 5) sin cos 0. 

13. Find the length of the curve of intersection of the 
elliptic cylinder a 2 y 2 + b 2 x 2 = a 2 b 2 , with the sphere 

x 2 + y 2 + z 2 = a 2 . 



CHAPTER VII. 



. AREAS OF PLANE CURVES. 

184. Areas of Curves.— Let PM and QN be two con- 
secutive ordinates of the curve AB, and let (x, y) be the 
point P ; let A denote the area included v 
between the curve, the axis of x, and 
two ordinates at a finite distance apart. 
Then the area of the trapezoid MPQN 
is an infinitesimal element whose breadth 
is dx and whose parallel sides are y and 
y + dy ; therefore we have 

dA = y + (y + *y) dz = ydx> 



Frg. 49. 



since the last term, being a differential of the second 
order, must be dropped. 

.*. A = / y dx, 

the integration being taken within proper limits. If, for 
example, we want the area between the two ordinates 
whose abscissas are a and b, where a > b, we have 






(1) 



In like manner, if the area were included between the 
curve, the axis of y, and two abscissas at a finite distance 
apart, we would have 



A = J xdy, 
where c and d are the ^-limits. 



(2) 



QUADRA TUBE OF THE CIRCLE. 



361 




185. Area between Two Curves. — If the area were 
included between the two curves AB and ab, whose equa- 
tions are respectively y = f (x) and y 
y = (p {x), and two ordinates CD and 
EH, where OD = b and OH == a, 
we should find by a similar course of 
reasoning, 

A = / [/ ( x ) — ( x )] dx. Fig. 50 

The determination of the area of a curve is called its 
Quadrature. 

186. The Circle. — The equation of the circle referred to 
its centre as origin, is y 2 = a 2 — x 2 ; therefore the area of 
a quadrant is represented by 

A = Art 2 - af)* dx 

rx (a 2 — x 2 )* a 2 • , %~\ a /« -^ . * *„,.% 

= — — 5 ^- + ^ snr 1 - (See Ex. 4, Art. 151.) 

L/C /« a J i) 



a 4 n 



therefore the area of the circle = -a 2 . 

Also, if OM = x, the area of OBDM 
becomes 



A = f\a 2 - art* 



dx 



_ [x {a 2 — x 2 ) 



[ 



~ + 2 Sm " 



1?T 

a-lo 



a; (a 2 — £ 2 ) a a 2 . , a; 
— — sr — - + 5- sm" 1 -. 




Fig. 51. 



This result is also evident from geometric considerations 
16 



302 



QUADRATURE OF THE PARABOLA. 



X 1 

for the area of the triangle OMD = '-(a? — x 2 )*, and the 

area of the sector ODB = - sin -1 -• 

2 a 

Eemark. — The student will perceive that in integrating between 
the limits x = and x — a, we take in every elementary slice PQRN 
in the quadrant ADBO ; also integrating between the limits x == 
and x — x = OM, we take in every elementary slice between OB 
and MD.* 



187. The Parabola.— From y 2 — 2px, 

we have 

y = \/2px. 

Hence, for the area of the part OPM, 
we have 



A = V%p I x^dx = | V%p & ; t. e„ %xy. 



Y 

o( 


M 


1 ^v 
H ^ 


P 



Fig. 52. 



Therefore the area of the segments POP', 

cut off by a chord perpendicular to the axis, is f of the 

rectangle PHH'P'. 

188. The Cycloid. — From the equation 



x = r vers -1 - — a/2/ 
r 



■y 



ft 



we have 



dx = 



_ y-dy 



\/%ry — y 2 



tfdy 






\^2ry — y 2 



= %i\r* 



(See Ex. 6, Art. 151) = \ the area of the cycloid. Since 
integrating between the limits includes half the area of 
the figure. 



* The student should pay close attention in every case to the limits of the 
integration. 



b r a i 

A = - I (a 2 — x 2 )'* dx 



AREA BETWEEN PARABOLA AND CIRCLE. 363 

Therefore the whole area = ^r 2 , or three times the area 
of the generating circle.* 

189. The Ellipse. — The equation of the ellipse referred 
to its centre as origin, is 

therefore the area of a quadrant is represented by 

I 

'o 

= --*£ (See Art 186) = ipbn. 

Therefore the area of the entire ellipse is 'nab. 

190. The Area between the Parabola y 2 = ax 
and the Circle y 2 = 2 ax — x 2 . — These curves pass 
through the origin, and also intersect at Y K 
the points A and B, whose common abscissa 
is a. Hence, to find the area included 
between the two curves on the positive side c 
of the axis of x, we must integrate between 
the limits x = and x = a. Therefore, 
by Art. 185, we have 



A = f\{2ax - x 2 )? — (ax)*] da 




«- o 



Fig. 53, 



= — \ *a- 



fa 2 ; (See Ex. 6, Art. 151.) 



which is the area of OPAP'. 



* This quadrature was first discovered by Roberval, one of the most distin- 
guished geometers of his day. Galileo, having failed in obtaining the quadrature 
by geometric methods, attempted to solve the problem by weighing the area of the 
curve against that of the generating circle, and arrived at the conclusion that the 
former area was nearly, but not exactly, three times the latter. About 16*8, 
Roberval attacked it, but failed to solve it. After studying the ancient Geometry 
for six years, he renewed the attack and effected a solution in 1634. (See Salmon's 
Higher Plane Curves, p. 266.) 



364 



THE SPIRAL OF ARCHIMEDES. 



191. Area in Polar Co-ordinates. — Let the curve be 
referred to polar co-ordinates, being the pole, and let 
OP and OQ be consecutive 
radii-vectores, and PR an 
arc of a circle described with 
as centre ; let (r, 6) be the 
point P.- Then the area of the 
infinitesimal element OPQ 
= OPE 4- PRQ ; but PRQ is 

an infinitesimal of the second order in comparison with 
OPR, when P and Q are infinitely near points ; conse- 




Fig. 54-. 



quently the elementary area OPQ = area OPR 



f*dO 

2 

Hence if A represents the area included between the curve, 
the radius-vector OP, and the radius-vector OB drawn to 
some fixed point B, we have 



= \frHQ. 



If (3 and a are the values of 6 corresponding to the points 
B and C respectively, we have 



f\m. 



192. The Spiral of Archimedes. — Let r = =~ be 

its equation. Then 

A = n frHr = -Jur 3 + C. 

If we estimate the area from the pole, we have A = 
when r = 0, and .*. (7=0; hence, 

A = %rrr*, 

which is the value of the area passed over by the radius- 
vector in its revolution from its starting at c to any value, 
as r. 



EXAMPLES. 365 

If we made d = 2n, we have r = 1 ; therefore 

A = ^ 

which is the area described by one revolution of the radius- 
vector. Hence the area of the first spire is equal to one- 
third the area of the measuring* circle. 
If we make = % (JJrr), r = % \ therefore 

A = f- 

which is the whole area described by the radius-vector 
during two revolutions, and evidently includes twice the 
first spire + the second. Hence the area of the first two 
spires = frr — Jt = %n, and so on. 

EXAMPLES. 

1. Find the area of y = x — x 3 between the curve and 
the axis of x. Ans. -J-. 

The limits will be found to be x = 0, x — + 1 ; also x — 0, 
x = - l.f 

2. -Find the area of y = x z — b 2 x between the curve 
and the axis of x. Ans. \¥. 

3. Find the area of y = x s — ax 2 between the curve and 
the axis of x. Ans. -f^aK 

4. Find the whole area of the two loops of a?y 2 
= x 2 (a 2 — x 2 ). Ans. %a 2 . 

5. Find the area of xy 2 = a? between the limits y = b 

and y = a . _ „b — c 

* Ans. 2a z —j — 

be 

6. Find the whole area of the two loops of aHf 
= a 2 b 2 x 2 — b 2 x\ xins. ^ab. 



* See Anal. Geom., Art 153. 

+ The student should draw the figure in every case, and determine the limits of 
the intentions. 



366 EXAMPLES. 

7. Find the whole area of a 2 y 2 = x 5 (2a — x).- (See 
Arts. 150 and 188.) Ans. -na 2 . 

8. Find the whole area between the Cissoid v 2 



V 



and its asymptote. (See Art. 103.) 2a ~ x 

J " v ; Ans. Sua 2 . 

9. The equation of the hyperbola is a 2 y 2 — b 2 x 2 = — a 2 b 2 ; 
find the area included between the curve, the axis of x, and 
an ordinate. X y ab . (x + \/x 2 — a 2 \ 



A xy ab , /c 



a / 

10. The equation of the Witch of Agnesi is 

x 2 y = 4ta 2 (2a — y) ; 

find the area included between the curve and its asymptote. 
(Use Formula 2 of Art. 184.) Ans. 4« 2 tt. 

11. Find the area of the catenary VPMO, Fig. 45. 



Ans. - le a '■ — e a J = a (y 2 — a 2 )K 



13. Find the area of the oval of the parabola of the third 
degree whose equation is cy 2 = (x — a) (x — b) 2 . (See 
Art. 142.) , 8 ,. .5 

13. Find the area of one loop of the curve 

ay 2 = x 2 (a 2 — x 2 )^. 

Ans. -fa 2 . 

14. Find the whole area between the curve 

x 2 y 2 + aW = a 2 y 2 
and its asymptotes. Ans. %-nab. 

15. Find the whole area of the curve 

&+&-■>■■ 

Ans. \-ndb. 



EXAMPLES. . 36 7 

16. Find the area included between the parabola if — 'Ipx 
and the right line y — ax. 

These two loci intersect at the origin and at the point whose ab- 
scissa is — ; hence the ^-limits are and — ; therefore, Art. 185, 
a 1 O" 

Pa* . 2r/ 

A t = J (\/2px — ax) dx — j- 3 , Arts. 

17. Find (2) the area included between the parabola 

y 2 = 2p%, 

the right line passing through the focus and inclined at 45° 
to the axis of x, and the left-hand double ordinate of inter- 
section. (See Art. 185.) 

Also find (2) the whole area between the line and parabola. 

(1.) Here the .r-limits are found tobe| (y 2 + l)" and | {*/%— if ; 
hence we have 

A= I [ V 2 P & dx-{x- ^p) dx] 



f(-tfM" 



So 



%\/%P x? — \x' 2 + ±px 

J f(t2-1) 2 

= W*P (I)* tt* + 2) - I (2 V2) \ + hp (4 V§) | 

= -W 2 - 3yV + W = f (-¥- - 2 a/2)> Ans. 
(2.) ' Ans. ip*\/2. 

18. Find the whole area included between the four infi- 
nite branches of the tractrix. Ans. rra 2 . 

19. Find the area of the Kaperian logarithmic spiral. 

Ans. Jr 2 . 

20. Find the whole area of the Lemniscate r 2 = a 2 cos 20. 

Ans. a 2 . 



368 EXAMPLES. 

21. Find the whole area of the curve 

r = a (cos 20 + sin 2(9). 

22. Find the area of the Cardioide. (See Art. 181.) 

Ans. ^-rra 2 . 

23. Find the area of a loop of the curve r = a cos nO. 

A *<& 

Ans. - — 

24. Find the area of a loop of the curve 

r = a cos nd + b sin nd. 

a? + b* n 

Ans. — 

4 n 

25. Find the area of the three loops of the curve 

r = a sin 3d. (See Fig. 33.) 

Ans. — — 
4 

26. Find the area included between the involute and the 
evolute in Fig. 46, when the string has made one revolution. 

Ans. fr 2 7r 3 . 



CHAPTER VIII 



AREAS OF CURVED SURFACES. 



193. Surfaces of Revolution. — If airy plane be sup- 
posed to revolve around a fixed line in it, every point in the 
plane will describe a circle, and any curve lying in the plane 
will generate a surface. Such a surface is called a surface 
of revolution ; and the fixed line, round which the revolu- 
tion takes place, is called the axis of revolution. 

Let P and Q be two consecutive 
points on the curve AB ; let (x, y) be 
the point P, and s the length of the 
curve AP measured from a fixed point 
A to any point P. Then MP = y, 
NQ = y-\-dy, and PQ == ds. 

Denote by S the area of the surface 
generated by the revolution of AP 
around the axis OX; then the surface generated by the 
revolution of PQ around the axis of x is an infinitesimal 
element of the whole surface, and is the convex surface of 
the frustum of a cone, the circumferences of whose bases 
are 2~y and 2n (y + dy), and whose slant height is PQ = ds ; 
therefore we have 




Fi ff . 55. 



dS 



2ny + 2n(y + dy) 

2 



PQ = 2-yds 



since the last term, being an infinitesimal of the second 
order, must be dropped. Therefore, for the whole surface, 
we have 

8 = 2tt fyds = 2- fyVdx* + dy\ 



370 QUADRATURE OF THE SPHERE, 

the integral being taken between proper limits. If for 
example, we want the surface generated by the curve be- 
tween the two ordinates whose abscissas are a and b, where 
a > b, we have 

In like manner it may be shown that to find the surface 
generated by revolving the curve round the axis of y, we 
have 

S = 2- Jxds. 

194. The Sphere. — From the equation of the gener- 
ating curve, x 2 -f y 2 = r 2 , we have 

y = (a* - x 2 )? and C -£ = - ? ; 



.-. S = 2n fy(l + T2 V dx = 2jt Jrdx = 2nrx + 



a 



Hence, the surface of the zone included between two 
planes corresponding to the abscissas a and b is 

S = 2ir Prdx — 2nr (a — b); 

that is, the area of the zone is the product of the circum- 
ference of a great circle by the height of the zone. 

To find the surface of the whole sphere, we integrate 
between + r and — r for the ^-limits ; hence we have 

S = 2nr f dx = 2nr [r — (— r)] = ±-r 2 ; 

that is, the whole surface of the sphere is four times the 
area of a great circle. 

Remark. — If a cylinder be circumscribed about a sphere, its convex 
surface is equal to 2~r x2r = 4~r-, which is the same as the surface 
of the sphere. If we add 2~?- 2 to this, which is the sum of the areas 
of the two bases, we shall have for the whole surface of the cylinder 



QUADRATURE OF FALIAB0L01D OF REVOLUTION. 371 

Q-r 2 . Hence the whole surface of the cylinder is to the surface of the 
sphere as 3 is to 2. This relation between the surfaces of these two 
bodies, and also the same relation between the volumes, was discovered 
by Archimedes, who thought so much of the discovery that he ex- 
pressed a wish to have for the device on his tombstone, a sphere 
inscribed in a cylinder. Archimedes was killed by the soldiers of 
Marcellus, b. c. 212, though contrary to the orders of that general. 
The great geometer was buried with honors by Marcellus, and the 
device of the sphere and cylinder was executed upon the tomb. 140 
years afterward, when Cicero was questor in Sicily, he found the 
monument of Archimedes, in the shape of a small pillar, and showed 
it to the Syracusans, who did not know it was in being ; he says it was 
marked with the figure of a sphere inscribed in a cylinder. The 
sepulchre was almost overrun with thorns and briars. See article 
" Marcellus," in Plutarch's Lives, Vol. Ill, p. 120. 

195. The Paraboloid of Revolution. — From the 
equation of the generating curve y 2 = 2jjx, we have 

y=V^x, and | = i \/|- 
- S = 2nf o X V2^(l + £fdz = 2WpfJ(p + ^dx 

= (V Vp (p + Ste)t] frr Vp l(p + &0* - P% (1) 

which is the surface generated by the revolution of the 
part of the parabola between its vertex and the point 
(x, y). 

We might have found the surface in terms of y instead 
of x, as follows : 

dx _ y 
dy ~ p 



2n py i 

= j J o v(p 2 + tf)*fy 



372 THE PROLATE SPHEROID. 

which result agrees with (1), as the student can easily 
verify. 

196. The Prolate Spheroid (See Anal. Geom., Art. 
191). — From the equation of the generating curve 

tf=(l — e 2 )(a 2 — x 2 ), 
we have 



2tt yds = 2tt Vl — e 2 / s/d 2 — x 2 ds 

= 2n Vl — e 2 Va 2 — e 2 x* dx (Art. 175.) 

(I ~ x2 f dx > 



, h la 2 

= 2rx -e 

a \ e* 



therefore for half the surface of the ellipsoid, since the 
^-limits are a and 0, we have 



a <Jo \e 2 J 

e-*) 



2ir_be 
a 



a 2 . A ex 



-f- — sm 



-i 



2 ' 2e 2 «Jo 

(See Ex. 4, Art. 151.) 

= w H sm * & 

197. The Catenary. — From the equation of the gen- 
erating curve, 

^ = 2 V + * 7* 



SURFACE GENERATED BY THE CYCLOID. 373 

we have for the surface of revolution around the axis of x 
between the limits x and 0, 

yds = na j (e« -f e «) ds 

— \~ a f X (e« + e~'a) dx (by Art. 177) 
rra P x ( 2 - --\ 

= YJo\ ea ^ 2 + e V dx 

fa 2 / - --\ 1 

= 7r (?/s -f- ax), (where s — VP, Fig. 45.) 

198. The Surface generated by the Cycloid when 
it revolves around its axis. — From its equation 



y = r vers -1 - + y 2rx — x 2 , (1) 



we have 



.-. S=2nfyds = Z*fy^dx. (4) 

Put u = y, dv = \/ — dx; 

.-. du = dy, and v = 2 V%rx; 
therefore (by Art. 147) we have 



374 POLAR CO-ORDINATES. 

j V \/ — d x — ty V%rx — 2 V%r I Vx dy 
= 2y V2rx — 2 V^J V2~r — x dx [by (2)] 

= 2 V%rz Ir vers -1 - -f V%rx — ~x^\ -f -f V%r (2r — %)% 

[by (1) and integrating.] 

■'■ £'*y/* *"=*"* -****, 

which in (4) gives 

S = SttV 2 — -^-rrr 2 == Snr 2 (n — |J. 

199. Surfaces of Revolution in Polar Co-ordi- 
nates. — If the surface is generated by a curve referred to 
polar co-ordinates, its area may be determined as follows : 
Let the axis of revolution be the initial line OX, see 
Fig. 54, and from P (r, 6) draw PM perpendicular to OX. 
Then PM = r sin 6, and the infinitesimal element PQ 
== ds will, in its revolution round OX, generate an infini- 
tesimal element of the whole surface, whose breadth = ds 
and whose circumference = 2nr sin 0. Hence, 

8 = /W sin $ ds* — 2n j V sin d (r 2 + S)^ 

(Art. 179) 
the integral being taken between proper limits. 

200. The Oardioide. — From Art. 181, we have 

ds = a (2 + 2 cos 6)h dd = 2a cos - dd. 



* This expression might have been obtained at once by substituting in Art. 193, 
for y, its value r sin 6. 



DOUBLE INTEGRATION. 



For the surface of revolution of the whole curve about 
the initial line, we have n and for the limits of 6, there- 
fore we have 



8 



= / 2nr sin 6 ds 

/** 

= 4=na 2 / (1 -f cos 6) cos - sin 6 dd 



f* n B 6 
= lf)7r« 2 / cos 4 - sin - dd 

— ["_ sf. n( fi cos 5 ? [= zg-rra 2 . 

201. Any Curved Surfaces. — Double Integration. — 

Let (x, y, z) and (x -f dx, y -\- dy, z -f dz) be two consecu- 
tive points p and q on the sur- 
face. Through p let planes be 
drawn parallel to the two planes 
xz and yz ; also through q let 
two other planes be drawn par- 
allel respectively to the first. 
These planes will intercept an 
infinitesimal element pq of the 
curved surface, and the projec- 
tion of this element on the 
plane of xy will be the infini- 
tesimal rectangle PQ, which = 

Let S represent the required area of the whole surface. 
and dS the area of the infinitesimal element pq, and 
denote by a, (3, y, the direction angles* of the normal at 
p (x, y, z). Then, since the projection of dS on the 
plane of xy is the rectangle PQ = dx dy, we have by Anal. 
Geom., Art, 168, 

dx dy = dS cos y. (1) 




Fig. 56. 



* See Anal. Goom.. Art. 170. 



376 SURFACE OF A SPHERE. 

Similarly, if dS is projected on the planes yz and zx, 
we have 

dy dz = dS cos a ; (2) 

ffe tf# = tf# cos ]8. (3) 

Squaring (1), (2) and (3), and adding, and extracting 
the square root, we have 

dS = (dtfdtf + efyW + dWa*)* 
(since cos 2 a -f cos 2 j3 -f cos 2 y = 1, 

Anal. Geom., Art. 170). 

.-. # = f f(dx 2 dy 2 + % 2 <fe 2 + <fe^fa?)» 






the limits of the integration depending upon the portion 
of the surface considered. 

202. The Surface of the Eighth Part of a Sphere.— 

Let the surface represented in Fig. 56 be that of the 
octant of a sphere ; then being its centre, its equation is 



Hence, 



X 2 + y2 _|_ z 2 — a \ 

dz x dz y 

dx ~ z 9 dy z 

r* r> adx dy 

J J Vet 2 — x 2 — y 2 

Now since pq is the element of the surface, the effect 
of a ^/-integration, x being constant, will be to sum up 
all the elements similar to pq from H to /; that is, 
from y = to y = hi = y x = Va 2 — x 2 ; and the aggre- 



EXAMPLES. 3t j 

gate of these elements is the strip Upl. The effect of a 
subsequent ^-integration will be to sum all these elemental 
strips that are comprised in the surface of which GAB 
is the projection, and the limits of this latter integra- 
tion must be x = and x — OA = a. Therefore, 
we have 

pa riy x a fa fly 

Jo Jq 



S 



pa ny x 



Va? — x 2 — y 2 
y* a d.r ill) 



«^o V.Vi 3 — y 2 

1 a dx sin 



i y T 1 






~a 2 



EXAMPLES. 

1. Find the convex surface of a right circular cone, 
whose generating line is ay — - hx = 0. 

A ns. -nb Va 2 + b 2 . 

Remark. — It is evident that the projection of the convex sur- 
face of a right circular cone on the plane of its base, is equal 
to the base; hence it follows (Anal. Geom., Art. 168) that the 
convex surface of a right circular cone is equal to the area of its 
base multiplied by the secant of the angle between the slant 
height and the base. Thus, calling this angle a, we have in 
the above example, 

8 = -&2 gee a = nW ^-^+— = tt& J~tfTl\ 

o v 

which agrees with the answer. 

2. Find the area of the surface generated by the revolu- 
tion of a logarithmic curve, y = e x , about the axis of x 9 
between the ^-limits and y. 

Ans. n\y(l +y)i + log [_y + (1 + # 2 )*] f. 



378 EXAMPLES. 

3. Find the area of the surface generated by the revolu- 
tion of the cycloid (1) about its base, and (2) about the 
tangent at the extremity of the base. 

Ans. (1) ^na 2 ; (0) ^~a 2 . 

4. Find the area of the surface generated by the revolu- 
tion of the catenary about the axis of y, between the 
^-limits and x. Ans. 2n [xs — a{y — a)\ 

By Art. 177, s = ^ (<?« — e~« \ ; 

.*. S =: 2tt I ccds = 2n \xs— I sdx , 
from which we soon obtain the answer. 

5. Find the area of the surface of a spherical sector, the 
vertical angle being 2« and the radius of the sphere = r. 

Ans. 47rr 2 /sin ^1 • 

6. Find the area of the surface generated by the revolu- 
tion of a loop of the lemniscate about its axis, the equation 
being r 2 == a 2 cos 26. j_ nSm wa 2 (2 _ 2*). 

Here find rds — a?dO ; /. etc. 

7. Find the area of the surface generated by the revolu- 
tion of a- loop of the lemniscate about its axis, the equation 
being r 2 = a 2 sin 20. Ans. 2na 2 . 

8. A sphere is cut by a right circular cylinder, the radius 
of whose base is half that of the sphere, and one of whose 
edges passes through the centre of the sphere. Find the 
area of the surface of the sphere intercepted by the cylinder. 

Let the cylinder be perpendicular to the plane of xy ; 
then the equations of the cylinder and the sphere are 
respectively y 2 = ax — x 2 and x 2 + y 2 -f z 2 = a 2 . It is 
easily seen that the ^/-limits are and Vax — x 2 == «/„ and 
the z-limits are and a. Therefore, Art. 201, we have 



EXAMPLES. 379 



P a PVi adxdy 
Jo Jo yV — x 2 — y : 



i 



— a I sin _1 ^ ax 

= a sin -1 !--- — rcZ(#-f #) 
i/o \« + xl 

= a [(«+■*) sin-* ^f-A, 



«.? 



(Art. 147) 



Therefore, the whole surface = 2a 2 (tt — 2). (In Price's 
Calculus, Vol. II, p. 326, the answer to this example is 
a 2 (7i — 2), which is evidently only half of what it should 
be.) 

9. In the last example, find the area of the surface of the 
cylinder intercepted by the sphere. 

Eliminating y, we have z = Vet 2 — ax for the equation 
of the projection on the plane xz of the intersection of the 
sphere and the cylinder. Therefore the z-limits are and 
2j == yV — ax, and the ^-limits are and a ; hence, Art. 
201, we have 



(dy\ 2 (dy\ 
\dxl "** \dz) 



dx dz 



[dx 2 dif + dy 2 dz 2 + dz 2 dx 2 ~f = 

a dx dz 

= — - — for an element of the surface of the cvlinder. 

2v ax — x 2 

a pa r*z x C J XC J Z a % r>a d x 

•'• S = O / / -T — = IT / ~ = a > 

4VQ e/Q \f ax — x 2 ^^o x* 
therefore the whole area of the intercepted surface of the 
cylinder is 4a a . (See Gregory's Examples, p. 436.) 

10. The axes of two equal right circular cylinders inter- 
sect at right angles ; find the area of the one which is inter- 
cepted by the other. Ans. 8a 2 . 



380 EXAMPLES. 

Let the axes of the two cylinders be taken as the axes of y and 2, 
and let a = the radius of each cylinder.. Then the equations are 

x- + s? = « 2 , x- + y- = «-. 

11- A sphere is pierced perpendicularly to the plane of 
one of its great circles by two right cylinders, of which the 
diameters are equal to the radius of the sphere and the axes 
pass through the middle points of two radii that compose a 
diameter of this great circle. Find the surface of that por- 
tion of the sphere not included within the cylinders. 

Ans. Twice the square of the diameter of the sphere. 

12. Find the area of the surface generated by the revolu- 
tion of the tractrix round the axis of x. Ans. 2ncfi. 

13. If a right circular cone stand on an ellipse, show that 
the convex surface of the cone is 

1 (OA 4- OA') (OA-OA')* sin «, . 

where is the vertex of the cone, A and A' the extremities 
of the major axis of the ellipse, and a is the semi-angle of 
the cone at the vertex. (See Remark to Ex. 1.) 



CHAPTER IX. 

VOLUMES OF SOLIDS. 

203. Solids of Revolution.— Let the curve AB, Fig. 55. 
revolve round the axis of x, and let V denote the volume 
of the solid bounded by the surface generated by the curve 
and by two planes perpendicular to the axis of x, one 
through A and the other through P ; then as MP and XQ 
are consecutive ordinates, the volume generated by the 1 evo- 
lution of MPQX round the axis of x is an infinitesimal 
element of the whole volume, and is the frustum of a cone, 
the circumferences of whose bases are 2ny and 2- (y + dy), 
and whose altitude is MX = dx; therefore we have 



dV = 



y£+^ + dyr + *yly + dy) dx = £ fdx , 



by omitting infinitesimals of the second order. Hence, for 
the whole volume generated by the area between the two 
ordinates whose abscissas are a and b, where a > b, we 
have 

V— I Ttlfdx. 

In like manner, it may be shown that to find the volume 
generated by revolving the arc round the axis of y, we have 

V= TrfxMy. 

204. The Sphere. — Taking the origin at the centre of 

the sphere, we have if = a- — x 2 ; therefore we have 



/a 
(a 
a 



2 _ a-2) dx = 



(atx — i^ 3 ) 






a. -~ n 



for the whole volume of the sphere. 



382 VOLUME GENERATED BY CYCLOID. 

Cor. 1. — To find the volume of a spherical segment be- 
tween two parallel planes, let b and c represent the distances 
of these planes from the centre ; then we have 

V=n f\a 2 - x*) dx ==7r[a 2 (b-c)-i (£ 3 - c 3 )]. 

J c 

Cor. 2. — To find the volume of a spherical segment with 
one base, let h be the altitude of the segment; then b = a 
and c = a — 7i, and we have 

V=7zf a (a 2 — x 2 ) dx = n¥ (a—~). 
J a-h \ of 

Cor. 3. f ~cfi = § of na 2 x 2a =. § of the circumscribed 
cylinder. (See Art. 194, Remark.) 

205. The Volume generated by the Revolution of 
the Cycloid about its Base. 

Here dx = _j^g = ( Art 176 ) . - 

V2ry — y 2 

and integrating between the limits y = and y = 2r, we 
find for the whole volume 

V = 2n I 



o */2ry — if 



s*2r 



= %*ir r "T — -- - (by Ex. 6, Art. 151) 

3 J ^/2ry -y 2 ' 

— ij> r7r (3 r 2 n ) (^ Ex. 6, Art. 151) 
__ 5-^2^ 

We have 5ttV 3 = |tt (2r) 2 x 2nr. 

Hence, the volume generated by the revolution of 
the cycloid about its base is equal to jive-eighths the 
circumscribing cylinder. 



SOLIDS BOUXDED BY AX*Y CURVED SURFACE. 



383 



206. The Cissoid when it revolves round its 
Asymptote. — Here 0M = x, MP — y, 
0A = 2ft, MA = 2a — x, HD = dy ; 
hence an infinitesimal element of the 
whole volume is generated by the revo- 
lution of PQDH about AT, aud is 
represented by 77 (2a — x)~ dy. 

The equation of the Cissoid is 

X s 




Fig. 57. 



y 



2 — 



2a — x 



(3a — x) (2ax — z 2 H ■ 
•'• **~ (2ft -*? * m > 

hence, between the limits x = and x = 2a, we have 
V = 2tt /(2a--xjdy = 2- / (3ft— x) (2ax— z*)$ dx 

1 



2tt 



A 2 



dr 



2~% 3 
(by Ex. 6, Art. 151). 



207. Volume of Solids bounded by any Curved 
Surface. — Let (x, y, z) and 

(x + dx, y + cly, z + ch) be two 
consecutive points E and F 
within the space whose volume 
j g to be found. Through E 
pass three planes parallel to 
the co-ordinate planes xy, yz, 
and zx; also through F pass 
three planes parallel to the 
first. The solid included by 
these six planes is an infinitesi- 
mal rectangular parallelopipe- 
don, of which E and F are two opposite angles, and the 
volume is dxdy dz ; the aggregate of all these solids between 




Fig. 58. 



384 TRIPLE INTEGRATION. 

the limits assigned by the problem is the required volume. 
Hence, if V denote the required volume, we have 



V 



j j j dxdydz, 



the integral being taken between proper limits. 

In considering the effects of these successive integrations, 
let us suppose that we want the volume in Fig. 58 contained 
within the three co-ordinate planes. 

The effect of the ^-integration, x and y remaining con- 
stant, is the determination of the volume of an infinitesimal 
prismatic column, whose base is dxdy, and whose altitude 
is given by the equations of the bounding surfaces ; thus, in 
Fig. 58, if the equation of the surface is z=f(x,y), the 
limits of the ^-integration are/ (x, y) and 0, and the volume 
of the prismatic column whose height is ~Pp is f(x, y) dx dy ; 
hence the integral expressing the volume is now a double 
integral and of the form 



V = j J f(x,y) dxdy. 



If we now integrate with respect to y, x remaining con- 
stant, we sum up the prismatic columns which form the 
elemental slice Uplmq, contained between two planes per- 
pendicular to the axis of x, and at an infinitesimal distance 
{dx) apart. The limits of y are J J and 0, 12 being the y to 
the trace of the surface on the plane of xy, and which may 
therefore be found in terms of x by putting z = in the 
equation of the surface : or, if the volume is included be- 
tween two planes parallel to that of xz, and at distances y Q 
and y x from it, y and y x being constants, they are in that 
case the limits of y; in the same way we find the limits if 
the bounding surface is a cylinder whose generating lines 
are parallel to the axis of z. In each of these cases the 
result of the ^-integration is the volume of a slice included 
between two planes at an infinitesimal distance apart, the 
length of which, measured parallel to the axis of y, is a 



EXAMPLES. 385 

function of its distance from the plane of yz ; thus the limits 
of the ^/-integration may be functions of x 9 and we shall 
have 

V = fff(z, y) dx dx, = J'F(x) dx, 

where F{x) dx is the infinitesimal slice perpendicular to 
the axis of x at a distance x from the origin, and the sum 
of all such infiuitesimal slices taken between the assigned 
limits is the volume. Thus, if the volume in Fig. 58 be- 
tween the three co-ordinate planes is required, and OA = a, 
then the .r-limits are a and 0. If the volume contained 
between two planes at distances x and x x from the plane of 
yz is required, then the ./-limits are x and x x . 

EXAMPLES. 

1. The ellipsoid whose equation is 
t . t . t - i 

f ,2 "I" £2 + c8 - X ' 

(3»2 W 2 \i 

1 — -g — Vo ) 

and 0, which call z x and ; the limits of y are hi = 
1 — — 2 )~ and 0, which call y x and ; the z-liinits are a 
and 0. 

First integrate with respect to z, and we obtain the infini- 
tesimal prismatic column whose base is PQ, Fig. 58, and 
whose height is T*p. Then we integrate with respect to y, 
and obtain the sum of all the columns which form the 
elemental slice Kplmq. Then integrating with respect to x, 
we obtain the sum of all the slices included in the solid 
OABC. 



\ V=8 f a f Vx f Z 'dxdydz 



17 



386 EXAMPLES. 

= -sJo Jo &-&***'*» 



-£rr!<w-** + « 



2 _ y m + ^_ s i n -i *_ 



Vl 7 



= — s- / (a 2 — a; 2 ) <£c == 4:ra #c. 

2. The volume of a right elliptic cylinder whose axis 
coincides with the axis of x and whose altitude = a, the 
equation of the base being 

Here the z-limits are 7 (J 2 — y 2 )i and 0, which call z x and 

; the ^-limits are b and ; the .r-limits are a and 0. 
ft 



pa po nz x 

\ V- l I I dxdydz 

«/q t/o «/0 






^ • 1^1* 7 







4 «/ 

(See Price's Calculus, Vol. II, p. 356.) 

3. The volume of the solid cut from the cylinder 'x 2 + y 2 
= a 2 by the planes z — and 2 == # tan a. 

Here the z-limits are x tan « and 0, or z x and 0; the 
^-limits are (a z — x 2 )* and — (# 2 — x 2 )^, or y, and — y x ; 
the z-limits are a and 0. 



EXAMPLES. 387 

pa py, pz 1 

.-. V- / / / dxdydz 

na r*y l 

= / / (a tan «) <fo c/«/ 

= 2 tan « / a; (a 2 — # 2 )« <£z =z § # 3 tan «. 
4. The volume of the solid common to the ellipsoid 

/v2 nii #2 

~2 + % -f -, = 1 and the cylinder x 2 + y 2 = 6\ 

( x 2 ? ,2\X 

1 — j 2 J 

and 0, or z x and ; the limits of the ^-integration are 
(b 2 — y 2 )h and 0, or x x and 0; the ^/-limits are and b.* 

pb px x pz x 

.-. V = 8 / / dydx dz 



>& /»«, |~ <y2 y2~ 



J a Jo \_ a 2 



" dy dx 



a? -]0'-v')i 



8c {"> \xl , ay ,\* V J . _. x , 

8 v-wJ. 



+« s ( i -S) sin_i lf ^ 



* la this example, this order of integration is simpler than it would be to take 
it with respect to y and tben x. 



388 EXAMPLES. 

= %abc\\ {a 2 - b 2 )? + sin" 1 -1. 
La 2 v y aJ 

(See Mathematical Visitor, 1878, p. 26.) 

208. Mixed System of Co-ordinates.— Instead of 

dividing a solid into columns standing on rectangular 

bases, so that z dx dy is the 

volume of the infinitesimal 

column, it is sometimes more 

convenient to divide it into 

infinitesimal columns standing 

on the polar element of area -. gg 

abed = r dr dd, in which case 

the corresponding parallelopipedon is represented by 

zr dr dO, and the expression for V becomes 




V = J Jzrdrdd, 



taken between proper limits. From the equation of the 
surface, z must be expressed as a function of r and 6. 

EXAMPLES. 

1. Find the volume included between the plane z = 0, 
and the surfaces x 2 + y 2 = kaz and y 2 = 2cx — x 2 . 

X 2 I ffi y* f 

Here z = — : — — = — ; hence the z-limits are — - and 0. 
4a 4« 4a 

The equation of the circle y 2 = 2cx — x 2 , in polar co-or- 
dinates, is r = 2c cos 6 ; hence the r-limits are and 

2c cos 6, or and r x ; and the 0-limits are and -• 



Jo Jo 



4a 



— / " cos 4 6 dd = •-- Aw. (Ex. 4, Art. 157.) 
a J a a lb v 



o 
.8 « 



EXAMPLES. 389 

2. The axis of a right circular cylinder of radius b, 
passes through the centre, of a sphere of radius a, when 
ay b ; find the volume of the solid common to both 
surfaces.* 

Take the .centre of the sphere as origin, and the axis of 
the cylinder as the axis of z\ then the equations of the 
surfaces are x 2 + y 2 + z 2 = a 2 and x 2 -f y 2 = b 2 ; or, in 
terms of polar co-ordinates, the equation of the cylinder 
is r == b. 

Hence for the volume in the first octant, the z-limits are 
yV _ #2~_ y2 or <\/a 2 — r 2 and ; the r-limits are b 

and ; the 0-limits are - and 0. 

' zr dr eld 



"0 ^0 



= 8 / I r (a 2 - r 2 )? eld elr 
Jo t7 o 



b 

eld 



= y iy - (« 2 - »)»]. 

(See Gregory's Examples, p. 428.) 

209. The polar element of plane area is r elr eld (Art. 
208). Let this element revolve round the initial line 
through the angle 2n, it will generate a solid ring whose 
volume is 2nr sin dr elr eld, since 2nr sin d is the circumfer- 
ence of the circle described by the point (r, d). Let <j> 
denote the angle which the plane of the element in any 
position makes with the initial position of the plane ; 
then el<p is the angle which the plane in any position makes 

* This example, as well as the preceding one, might be integrated directly in 
terms of x and y by the method of Art. 207, but the operation would be more com- 
plex than the one adopted. 



390 EXAMPLES. 

with its consecutive position. The part of the solid ring 
which is intercepted between the revolving plane in these 
two consecutive positions, is to the whole ring in the same 
proportion as d<j> is to 2n. Hence the volume of this 
intercepted part is 

r 2 sin 6 d<p dd dr, 

which is therefore an expression in polar co-ordinates for 
an infinitesimal element of any solid. Hence, for the 
volume of the whole solid we have 

V — f j fr 2 sin d dj> dd dr, 

in which the limits of the integration must be so taken 
as to include all the elements of the proposed solid. In 
this formula r denotes the distance of any point from the 
origin, B denotes the angle which this distance makes with 
some fixed right line through the origin (the initial line), 
and denotes the angle which the plane passing through 
this distance and the initial line makes with some fixed 
plane passing through the initial line. (See Lacroix Oal- 
cul Integral, Vol. II, p. 209.) 

The order in which the integrations are to be effected is 
theoretically arbitrary, but in most cases the form of the 
equations of surfaces makes it most convenient to integrate 
first with respect to r ; but the order in which the 0- and 
^-integrations are effected is arbitrary. 

EXAMPLES. 

1. The volume of the octant of a sphere. Let a = the 
radius of the sphere ; then the limits of r are and a ; 
hence, 

V =. f ft sm OdfdO. 

In thus integrating with respect to r, we collect all the 
elements like r 2 sin d(f> dJQ dr which compose a pyramidal 



EXAMPLES. 391 

solid, having its vertex at the centre of the sphere, and for 
its hase the curvilinear element of spherical surface which 
is denoted by a 2 sin 6 d(f> dd. 

Integrating next with respect to between the limits 

and - , we have 

r = /| 3 [(- cose )]> = /f* 

In thus integrating with respect to 6, we collect all the 

pyramids similar to - sin 6 d<p dd, which form a wedge- 
o 

shaped slice of the solid contained between two consecutive 
planes through the initial line. 

71 

Lastly, integrating with respect to from to -, 
we have 

V = -|- • (See Todhunters Int. CaL, p. 183.) 

In this example the order of the integrations is imma- 
terial. 

2. The volume of the solid common to a sphere of 
radius a, and the right circular cone whose vertical angle 
is 2« and whose vertex is at the centre of the sphere. 

Here the r-limits are and a, the 0-limits are and a, 
the 0-limits are and 2^. 

.-. V = / / / r°~ sin 6 d(p dd dr 
sin 6 d<p dd 



ffl 



v 

ft'lTT C( $ 

/ r- (1 — COS «) d<p 

*J o 

|t« 3 (1 — COS a). 



392 EXAMPLES. 



EXAMPLES. 

1. Find the volume of a paraboloid of revolution whose 
altitude = a and the radius of whose base = b. 

71 

Ans. -x ab 2 . 

2. Find the volume of the prolate spheroid. Also of 
the oblate spheroid. Ans. The prolate spheroid = f nab 2 . 

The oblate spheroid = ^a 2 b. 

3. Find the volume of the solid generated by the revolu- 
tion of y = a x about the axis of x 9 between the limits x 

and — oo, where a > 1. n ._, 

Ans. - a 2x (log a) *. 

4. Find the volume of the solid generated by the revolu- 
tion of y = a log x about the axis of x, between the 
limits x and 0. Ans. ^a 2 x (log 2 x — 2 log x -f 2). 

5. Find the volume of the solid generated by the 
revolution of the tractrix round the axis of x. Ans. § *a % . 

6. Find the volume of the solid generated by the 
revolution of the catenary round the axis of x. 

Ans. - a (ys -f- ax). (Compare with Art. 197.) 

1. Find the volume generated by the revolution of a 
parabola about its base 2b, the height being h.* (See 
Art. 206.) Ans. \^bh\ 

8. The equation of the Witch of Agnesi being 

2a — tf 



• 



4,a 2 



y 



find the volume of the solid generated by its revolution 
round the asymptote. Ans. 4 A 3 . 



* This solid is called a parabolic spiudle. 



EXAMPLES. S93 

9. Find the volume of a rectangular parallelopipedon, 
three of whose edges meeting at a point are a, h, c. (See 
Art. 207.) Ans. abc. 

10. Find the volume contained within the surface of an 
elliptic paraboloid * whose equation is 

?/ 2 Z 2 

a + h = 2X > 

a o 
and a plane parallel to that of yz, and at a distance c from it. 

Ans. -c z (ab)*. 

11. The axes of two equal right circular cylinders inter- 
sect at right angles, their equations being x 2 -f z 2 = a? and 
x 2 -f y 2 = a 2 ; find the volume of the solid common to both. 

Ans. ^-a 3 . 

12. A paraboloid of revolution is pierced by a right cir- 
cular cylinder, the axis of which passes through the focus 
and cuts the axis of the paraboloid at right angles, the 
radins of the cylinder being one-fourth the latus-rectum of 
the generating parabola ; find the volume of the solid com- 

Here the equations of the surfaces are 

g/ 2 + 2- = 2px and .r 2 + y' 2 = px. 

13. Find the volume of the solid cut from the cylinder 

x 2 + y 2 = 2ax by the planes z = x tan « and z — x tan (3. 

-a 3 
Ans. 2 (tan (3 — tan a) -»- • 

14. Find the volume of the solid common to both sur- 
faces in Ex. 8 of Art. 202. (See Art. 208.) 

Ans. t(3:r — 4) a 3 . 

15. Find the volume of the part of the hemisphere in the 
last example, which is not comprised in the cylinder. 

Ans. fa 3 . 

* Called elliptic paraboloid because the sections made by planes parallel to the 
planes of xy and xz are parabolas, while those narallel to the plane of yz are 
ellipses. (Salmon's Anal. Geom. of Three Dimensions, p. 58.) 



mon to the two surfaces. „ /2 

Ans 



394 EXAMPLES. 

16. Find the volume of the solid intercepted between the 
concave surface of the sphere and the convex surface of the 
cylinder in Art. 208, Ex. 2. j^s. &n (a 2 — b 2 )?. 

17. Find the volume of the solid comprised between the 

surface z = ae c * and the plane of xy. Ans. rrac 2 . 

Here the r-limits are and oo ; and the 0-limits are and 2~. 

18. Find the volume of the solid generated by the revo- 
lution of the cardioide r = a (1 + cos 6) about the initial 
line. 

pn P'Utt /?a(l + cos0) 

Here V = / / / r 2 sin 6 dd cfy dr = etc. 

e/Q *SQ t/0 

(See Art. 209.) Sna s 

Jtns. — ^ — • 

o 

19. Find the volume of the solid generated by the revo- 
lution of the Spiral of Archimedes, r = ad, about the initial 
line between the limits 6 = tt and 6 = 0. 

Ans. |-%3( /r 2_ e), 

20. A right circular cone w T hose vertical angle = 2«, has 

its vertex on the surface of a sphere of radius a, and its 

axis coincident with the diameter of the sphere ; find the 

volume common to the cone and the sphere. 

. 4uft 3 _ . 

Ans. — — (1 — cos 4 a). 

o 

21. Find the volume of a chip cut at an angle of 45° to 
the centre of a round log with radius r. (Mathematical 
Visitor, 1880, p. 100.) Ans. |r 3 . 

22. Find the volume bounded bv the surface 

©'+ ©'+©' = ' 

and the positive sides of the three co-ordinate planes. 

abc 
Ans. w 



EXAMPLES. 395 

23. Find the volume of the solid bounded by the three 
surfaces x 2 + y 2 = cz, x 2 + y 2 = ax, and z = 0. 

3~^ 

*"■ «r 

24. A paraboloid of revolution and a right circular cone 
have the same base, axis, and vertex, and a sphere is 
described upon this axis as diameter. Show that the volume 
intercepted between the paraboloid and cone bears the same 
ratio to the volume of the sphere that the latus-rectum of 
the parabola bears to the diameter of the sphere. 

25. Find the volume included between a right circular 
cone whose vertical angle is 60° and a sphere of radius r 
touching it along a circle, by the formula 

V = / / I dx dy dz, 

. nr 3 
Am. T . 

26. In the right circular cone given in Ex. 13 of Art. 
202., prove that its volume is represented by 

^(OA.OA')*sin2«cos«. 









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